Law of Dynamics rotary motion solid body in the projection on the axis of rotation z: , Where I z is the moment of inertia of the body relative to axes of rotation, is the projection of the angular acceleration onto the axis of rotation, is the sum of the projections of the external moments of forces, is the projection of the angular momentum of the rigid body.
where is the radius vector of the force application point . , , are projections of the moment of force. Modulus of moment of force 
or , where a is the angle between the force and the radius vector .
7-1. Thin homogeneous mass rod m= 1 kg and length l= 1 m can rotate in a vertical plane around a horizontal axis passing through its end. The moment of friction forces M tr acts in the axis. = 1 N×m. The rod is brought to a horizontal position and released without a push. Find the angular acceleration at the initial moment of time. g \u003d 10 m / s 2.
7-2. Thin homogeneous mass rod m and length l can rotate in a vertical plane without friction about a horizontal axis passing through its end. The rod is placed at an angle a to the horizontal and released without a push. Find its angular acceleration at the initial time. m= 1 kg, l\u003d 1 m, a \u003d 30 °, g \u003d 10 m / s 2.
7-3. Thin homogeneous mass rod m= 1 kg length l= 1 m can rotate in a horizontal plane without friction around a vertical axis WITH passing through the middle of the rod. To the end of the rod in the plane of rotation at an angle a = 30°, a force = 1 N is applied to the rod. Find the angular acceleration of the rod at the initial moment of time.
7-4. Thin homogeneous mass rod m and length l can rotate in a horizontal plane around a vertical axis WITH passing through the middle of the rod. The moment of the friction force M tr acts in the axis. A force is applied to the end of the rod in the plane of rotation perpendicular to the rod. Find the angular acceleration of the rod at the initial moment of time.
m= 1 kg, l= 1 m, F\u003d 3 N, M tr \u003d 1 N × m.
7-5. A thin homogeneous plate in the form of a square with a side b WITH WITH equals I. A small weight of mass was glued to the middle of the side of the square. m and let go without a hitch. At the initial moment, the side of the square was vertical. Find the angular acceleration of the resulting figure at the initial moment of time. m= 1 kg, I= 1 , b\u003d 1 m, g \u003d 10 m / s 2.
7-6. A thin homogeneous rectangular plate with sides b And a can rotate without friction in a vertical plane around a horizontal axis passing through the center of mass WITH. Moment of inertia of the plate about the axis WITH equals I. A small weight of mass was glued to the middle of the side of the plate. m and let go without a hitch. At the initial moment, the side of the plate was vertical. Find the angular acceleration of the resulting figure at the initial moment of time.
m= 1 kg, I= 1 , b= 1 m, a\u003d 2 m, g \u003d 10 m / s 2.
7-7. Some body rotates around a fixed axis without friction. Its angular momentum relative to the axis of rotation depends on time according to the law. Through time t=1 s the body has an angular acceleration e. Find the moment of inertia of the body if t = 1 s. A\u003d 1, e \u003d 1 rad / s 2.
7-8. The body rotates around a fixed axis with angular acceleration, the time dependence of which is given by the graph. The moment of inertia of the body about the axis of rotation is I. Find the angular momentum of the body at time s, if s is -2. I= 1
3.1. Find moment of inertia J and moment of momentum L the globe about the axis of rotation.
3.2. Two balls of the same radius R = 5 cm are fixed at the ends of a weightless rod. The distance between the balls is r = 0.5 m. The mass of each ball is m = 1 kg. Find: a) the moment of inertia Jl of the system about the axis passing through the middle of the rod perpendicular to it; b) the moment of inertia J2 of the system about the same axis, counting the balls material points, whose masses are concentrated at their centers; c) the relative error S = (J1 - J2)/J2, which we allow when calculating the moment of inertia of the system, replacing the value of J1 with the value of J2.
3.3. A tangential force F = 98.1N is applied to the rim of a homogeneous disk with a radius R = 0.2 m. During rotation, the moment of friction forces acts on the disk Mtr = 98.1 N m. Find the mass m of the disks if it is known that the disk rotates with an angular acceleration e = 100 rad/s2.
3.4. A homogeneous rod with a length L = 1 m and a mass m - 0.5 kg rotates in a vertical plane around a horizontal axis passing through the middle of the rod. With what angular acceleration e does the rod rotate if the moment of forces M = 98.1 mN m acts on it?
3.5. A homogeneous disk with a radius R = 0.2 m and a mass m = 0.5 kg rotates around an axis passing through its center perpendicular to its plane. The dependence of the angular velocity of disk rotation on time t is given by the equation to = A + Bt, where B = 8 rad/s2. Find the tangential force F applied to the disk rim. Ignore friction.
The law of dynamics of rotational motion of a rigid body in projection onto the axis of rotation z: , Where I z is the moment of inertia of the body about the axis of rotation, is the projection of the angular acceleration onto the axis of rotation, is the sum of the projections of external moments of forces, is the projection of the angular momentum of the rigid body.
where is the radius vector of the force application point . , , are projections of the moment of force. Modulus of moment of force 
or , where a is the angle between the force and the radius vector .
7-1. Thin homogeneous mass rod m= 1 kg and length l= 1 m can rotate in a vertical plane around a horizontal axis passing through its end. The moment of friction forces M tr acts in the axis. = 1 N×m. The rod is brought to a horizontal position and released without a push. Find the angular acceleration at the initial moment of time. g \u003d 10 m / s 2.
7-2. Thin homogeneous mass rod m and length l can rotate in a vertical plane without friction about a horizontal axis passing through its end. The rod is placed at an angle a to the horizontal and released without a push. Find its angular acceleration at the initial time. m= 1 kg, l\u003d 1 m, a \u003d 30 °, g \u003d 10 m / s 2.
7-3. Thin homogeneous mass rod m= 1 kg length l= 1 m can rotate in a horizontal plane without friction around a vertical axis WITH passing through the middle of the rod. To the end of the rod in the plane of rotation at an angle a = 30°, a force = 1 N is applied to the rod. Find the angular acceleration of the rod at the initial moment of time.
7-4. Thin homogeneous mass rod m and length l can rotate in a horizontal plane around a vertical axis WITH passing through the middle of the rod. The moment of the friction force M tr acts in the axis. A force is applied to the end of the rod in the plane of rotation perpendicular to the rod. Find the angular acceleration of the rod at the initial moment of time.
m= 1 kg, l= 1 m, F\u003d 3 N, M tr \u003d 1 N × m.
7-5. A thin homogeneous plate in the form of a square with a side b WITH WITH equals I. A small weight of mass was glued to the middle of the side of the square. m and let go without a hitch. At the initial moment, the side of the square was vertical. Find the angular acceleration of the resulting figure at the initial moment of time. m= 1 kg, I= 1 , b\u003d 1 m, g \u003d 10 m / s 2.
7-6. A thin homogeneous rectangular plate with sides b And a can rotate without friction in a vertical plane around a horizontal axis passing through the center of mass WITH. Moment of inertia of the plate about the axis WITH equals I. A small weight of mass was glued to the middle of the side of the plate. m and let go without a hitch. At the initial moment, the side of the plate was vertical. Find the angular acceleration of the resulting figure at the initial moment of time.
m= 1 kg, I= 1 , b= 1 m, a\u003d 2 m, g \u003d 10 m / s 2.
7-7. Some body rotates around a fixed axis without friction. Its angular momentum relative to the axis of rotation depends on time according to the law. Through time t=1 s the body has an angular acceleration e. Find the moment of inertia of the body if t = 1 s. A\u003d 1, e \u003d 1 rad / s 2.
7-8. The body rotates around a fixed axis with angular acceleration, the time dependence of which is given by the graph. The moment of inertia of the body about the axis of rotation is I. Find the angular momentum of the body at time s, if s is -2. I= 1
The law of dynamics of rotational motion of a rigid body in projection onto the axis of rotation z: , Where I z
is the moment of inertia of the body about the axis of rotation, 
is the projection of the angular acceleration onto the axis of rotation, 
is the sum of projections of external moments of forces, 
is the projection of the angular momentum of the rigid body.
,
Where 
is the radius vector of the force application point 
.
,
,
are the projections of the moment of force. Modulus of moment of force 
or 
, where is the angle between the force 
and radius vector 
.
6-1. Thin homogeneous mass rod m
= 1 kg and length l= 1 m can rotate in a vertical plane around a horizontal axis passing through its end. The moment of friction forces M tr acts in the axis. = 1 Nm. The rod is brought to a horizontal position and released without a push. Find the angular acceleration at the initial moment of time. g \u003d 10 m / s 2.
Answer: 12 rad/s 2
6-2. Thin homogeneous mass rod m and length l can rotate in a vertical plane without friction about a horizontal axis passing through its end. The rod is placed a) at an angle to the horizon;
b) at an angle to the vertical and released without a push. Find its angular acceleration at the initial time. m = 1 kg, l \u003d 1 m, \u003d 30, g \u003d 10 m / s 2.
Answers: a) 13 rad / s 2; b) 7.5 rad/s 2
6-3. Thin homogeneous mass rod m= 1 kg
and length l
= 1 m can rotate in a horizontal plane without friction around a vertical axis WITH passing through the middle of the rod. To the end of the rod in the plane of rotation at an angle = 30, a force is applied to the rod 
=1 N. Find the angular acceleration of the rod at the initial moment of time.
Answer: 3 rad/s 2
6-4. Thin homogeneous mass rod m
and length l
can rotate in a horizontal plane around a vertical axis WITH passing through the middle of the rod. The moment of the friction force M tr acts in the axis. A force is applied to the end of the rod in the plane of rotation perpendicular to the rod 
. Find the angular acceleration of the rod at the initial moment of time.
m = 1 kg, l = 1 m, F= 3 N, M tr = 1 Nm.
Answer: 6 rad/s 2
6-5. A thin homogeneous plate in the form of a square with a side b
WITH WITH equals I. A small weight of mass was glued to the middle of the side of the square. m and let go without a hitch. At the initial moment, the side of the square was vertical. Find the angular acceleration of the resulting figure at the initial moment of time. m
= 1 kg, I
=
1
,b\u003d 1 m, g \u003d 10 m / s 2.
Answer: 4 rad/s 2
6-6. A thin homogeneous rectangular plate with sides b And a can rotate without friction in a vertical plane around a horizontal axis passing through the center of mass WITH. Moment of inertia of the plate about the axis WITH equals I. A small weight of mass was glued to the middle of the side of the plate. m and let go without a hitch. At the initial moment, the side of the plate was vertical. Find the angular acceleration of the resulting figure at the initial moment of time.
m= 1 kg, I= 1
,
b= 1 m, a\u003d 2 m, g \u003d 10 m / s 2.
Answer: 5 rad/s 2
6-7. Thin uniform length rod l
can rotate in a horizontal plane around a vertical axis passing through the middle of the rod. Force applied to the end of the rod 
. What is the projection of the moment of force relative to the point WITH per axle z.
l= 1 m, A = 1 N, IN= 2 N, D= 3 N. Answer: -0.5 Nm
6-8. A small ball is placed at a point with a radius vector 
A) 
; b) 
; V) 
. Find the modulus of the moment of force relative to the origin.
A = 1 m, IN= 2 m, WITH= 3 m, D= 4 N, .
Answers: a) 14.42 Nm; b) 12.65 Nm; c) 8.94 Nm
6-9. A small ball is placed at a point with a radius vector 
. At some point, the ball was acted upon by a force 
. Find the projection of the moment of force relative to the origin a) on the axis X; b)
per axle y; c) on the axle z
A = 1 m, IN= 2 m, WITH= 3 m, D= 3 N, E= 4 N, G= 5 N.
Answers: a) -2 Nm; b) 4 Nm; c) –2 Nm
6-10. Some body rotates around a fixed axis without friction. Its angular momentum relative to the axis of rotation depends on time according to the law
A) 
; b) 
; V) 
; G) 
; e) 
. Through time t\u003d 1 s the body has an angular acceleration . Find the moment of inertia of the body if =1 s. A
= 1
, \u003d 1 rad / s 2.
Answers: a) 1 kgm 2; b) 2 kgkg 2; c) 3 kgm 2; d) 4 kgm 2; e) 5 kgm 2
6-11. The body rotates around a fixed axis with angular acceleration, the time dependence of which is given by the graph. The moment of inertia of the body about the axis of rotation is I.
Find the angular momentum of the body at a moment in time 
with if 
s -2 . I
= 1
Answer: 1 Nms
6-12. The body rotates around a fixed axis with an angular velocity, the time dependence of which is given by a graph. The moment of inertia of the body about the axis of rotation is I. Find
a) the ratio of the modules of the moments of forces;
b) how much the modules of the moments of forces differ,
acting on the body at times 
with and 
With. 
s –1 , I
=
1
Answers: a) 0.5; b) 0.5 
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3.1. Find moment of inertia J and angular momentum L of the globe about the axis of rotation.
Solution:
3.2. Two balls of the same radius R = 5 cm are fixed at the ends of a weightless rod. Distance between balls r = 0.5 m. Mass of each ball m= 1 kg. Find: a) moment of inertia J 1 system with respect to the axis passing through the middle of the rod perpendicular to it; b) the moment of inertia J 2 of the system about the same axis, considering the balls as material points, the masses of which are concentrated at their centers; c) relative error b = (J 1 - J 2) / J 2 , which we allow when calculating the moment of inertia of the system, replacing the quantity J 1 magnitude J 2 .
Solution:
3.3. To the rim of a homogeneous disk with a radius R = 0.2 m applied tangential force F \u003d 98.1 N. During rotation, the friction torque M tr acts on the disk = 98.1 Nm . Find mass m disks, if it is known that the disk rotates with angular acceleration e\u003d 100 rad / s 2.
Solution:
3.4. Homogeneous rod with length l = 1 m and mass m= 0.5 kg rotates in a vertical plane around a horizontal axis passing through the middle of the rod. With what angular acceleration e a rod rotates if a moment of force acts on it M= 98.1 mN*m?
Solution:
3.5. Homogeneous disk with radius R = 0.2 m and weight m= 0.5 kg rotates around an axis passing through its center perpendicular to its plane. Angular velocity dependence w disk rotation versus time t is given by the equation w= A +bt, Where B = 8 rad/s 2 . Find the tangential force F, attached to the rim of the disk. Ignore friction.
Solution:
3.6. Flywheel whose moment of inertia J = 63.6kgm 2 rotation with angular velocity w = 31.4 rad/s. Find the moment of forces torus M, under the action of which the flywheel stops after a time t = 20 s. The flywheel is considered to be a homogeneous disk.
Solution:
3.7. To the wheel rim with a radius of 0.5 m and a mass m = 50 kg at tangential force F = 98.1 N. Find the angular acceleration s wheels. After what time t after the start of the force, the wheel will have a speed of rotation n= 100 rpm? The wheel is considered to be a homogeneous disk. Ignore friction.
Solution:
3.8. Handwheel radius R = 0.2 m and weight m = 10 kg is connected to the motor with a drive belt. The tension force of the belt running without slipping, T = 14.7N. What frequency vra n will have a flywheel after a time t = 10 s after the start of movement? The flywheel is considered to be a homogeneous disk. Ignore friction.
Solution:
3.9. A flywheel whose moment of inertia is J = 245 kg l, rotates with a frequency n= 20 rpm. After a time t = 1 min after the moment of forces ceased to act on the wheel M, it stopped. Find the moment of friction forces and the number of revolutions N, which made the wheel to a complete stop after the cessation of the forces. The wheel is considered to be a homogeneous disk.
Solution:
Z.10. Two weights with masses m 1 =2 kg and m2\u003d 1 kg are connected by a thread thrown over a block with a mass m= 1 kg. Find acceleration A, with which the weights move, and the tension forces T 1 and T 2 threads to which weights are suspended. The block is considered to be a homogeneous disk. Ignore friction.
Solution:
3.11. A cord is wound on a drum of mass m 0 \u003d 9 kg, to the end of which a load of mass m = 2 kg. Find acceleration A group Consider the drum as a homogeneous cylinder. Friction prene.
Solution:
3.12. On drum radius R = 0.5 m cord is wound, to the end of which is tied a load of mass m= 10 kg. Find moment of inertia J drum, if it is known that the load is descending with acceleration a = 2.04 m/s 2 .
Solution:
3.13. On drum radius R = 20 cm, moment of inertia koto J = 0.1 kgm 2, a cord is wound, to the end of which a load of mass m= 0.5 kg. Before the drum begins to rotate, the height of the load above the floor h Q = 1 m. After what time t will the load drop to the floor? Find kinetic energy W K load at the moment of impact on the floor and the tension force of the thread T. Ignore friction.
Solution:
3.14. Two weights with different masses are connected by a thread, passing through a block, the moment of inertia of which is J = 50 kgm 2 and radius R = 20 cm. Rotating block friction torque = 98.1 Nm. Find the difference in the tension forces of the thread T1-T2 on both sides of the block, if the block is known to rotate with an angular acceleration e \u003d 2.36 rad / s 2. The block is considered to be a homogeneous disk.
Solution:
3.15. Mass block m= 1 kg is fixed at the end of the table (see Fig. and problem 2.31). Weights 1 and 2 of the same mass m 1 =m 2 =1kg are connected by a thread thrown over the block. Coefficient of friction weight 2 on the table To= 0.1 . Find acceleration A, with which the weights move, and the tension forces T1 And T2 threads. The block is considered to be a homogeneous disk. Ignore friction in the block.
Solution:
3.16. Disk mass m \u003d 2 kg rolls without slipping along the horizontal plane with a speed v \u003d 4 m / s. Find the throw energy W k disk.
Solution:
3.17. Ball diameter D = 6 cm and mass m = 0.25 kg rolls without slipping on a horizontal plane with a rotation frequency n= 4 rpm. Find kinetic energy W K ball.
Solution:
3.18. Hoop and disk of the same mass m 1 = m 2 roll without sliding with the same speed v. Kinetic hoop energy W Kl =4kgcm. Find kinetic energy W k2 disk.
Solution:
3.19. Ball mass m= 1 kg rolls without slipping, hits the wall and rolls away from it. Ball speed before hitting the wall v = 10 cm/s, after hitting u= 8 cm/s. Find the amount of heat Q released when the ball hits the wall.
Solution:
3.20. Find the relative error b, which will be obtained when calculating kinetic energy W K rolling ball, if the rotation of the ball is not taken into account.
Solution:
