As shown in Figure 2, a disk rotates at a uniform speed around the vertical axis passing through the center in the horizontal plane, and a small object on the disk is stationary relative to the disk and moves with the disk. The centripetal force on the object is correct in the following statement: a A. The direction of the centripetal force points to the center of the disc B. the direction of the centripetal force is the same as the speed direction of the object C. The centripetal force direction is parallel to the rotating shaft D. the centripetal force direction remains unchanged The disc rotates counterclockwise

As shown in Figure 2, a disk rotates at a uniform speed around the vertical axis passing through the center in the horizontal plane, and a small object on the disk is stationary relative to the disk and moves with the disk. The centripetal force on the object is correct in the following statement: a A. The direction of the centripetal force points to the center of the disc B. the direction of the centripetal force is the same as the speed direction of the object C. The centripetal force direction is parallel to the rotating shaft D. the centripetal force direction remains unchanged The disc rotates counterclockwise

How do you explain that?
It is said in the textbook that the centripetal force always points to the center of the circle, and the centripetal force is always perpendicular to the speed
I don't need any explanation

As shown in the figure, there is a horizontal disc that can rotate around the vertical central axis, on which a spring with stiffness coefficient K is placed. One end of the spring is fixed on the shaft o, and the other end is connected with a small block a (which can be regarded as a particle) with mass M. the dynamic friction coefficient between the block and the disc is μ， At the beginning, the spring does not deform, and the length is l0. If the maximum static friction is equal to the sliding friction, the gravitational acceleration is g, and the object block a always rotates with the disc. Then: (1) What is the angular velocity of the disc at which block a will begin to slide? (2) When the angular velocity of the disc slowly increases to 4 μ g What is the elongation of the spring at l0? (the extension of the spring is within the elastic limit and the block is not separated from the disc)

(1) Let the angular velocity of the disc be ω At 0, block a will begin to slide, then μ mg=mR ω 02:00     Solution   ω 0＝ μ Gl0 (2) set the elongation of the spring at this time as △ x, then μ mg+k△x=mr ω 2, r = R + △ x, the solution is    △x＝15 μ mgL0kL0−16 μ mg...

What is the moment of inertia of a homogeneous disk with mass m and radius r about an axis passing through the center of the disk and perpendicular to the disk surface? If the rotational angular velocity is ω, Then its effect on the rotating shaft

mr ²/ 2 moment of momentum WMR ²/ two

As shown in the figure, put a 1kg wooden block on the horizontal ground, and the dynamic friction coefficient between the wooden block and the ground is 0.6. Simultaneously apply two mutually perpendicular tensile forces F1 and F2 to the wooden block in the horizontal direction. It is known that F1 = 3N and F2 = 4N. If the maximum static friction force is equal to the sliding friction force, G is taken as 10N / kg, then (1) What is the friction force on the block? (2) If F2 is turned 90 ° clockwise, what is the resultant force on the wood block?

(1) It can be seen from the figure that according to the parallelogram rule of force, the resultant force F of two tensile forces can be obtained=
32+42 N=5N，
The maximum static friction of wood block, i.e. sliding friction, is f= μ N=0.6 × 10N = 6N, so it can't be pulled,
According to the force balance conditions, the friction force on the wood block is equal to the resultant force of the tensile force, that is, 5N;
(2) When F2 is turned 90 ° clockwise, the resultant force of the two pulling forces is 7n, so the resultant force of the wood block in the horizontal direction is equal to the difference between the resultant force of the two pulling forces and the sliding friction, that is, f = 7n-6n = 1n
Answer: (1) the friction force on the wood block is 5N;
(2) If F2 is turned 90 ° clockwise, the resultant force on the wood block is 1n

As shown in the figure, the block with mass m = 1kg is placed at an inclination of θ On the slope of, the slope mass m = 2kg, and the dynamic friction coefficient between the slope and the wood block is 0.2 The ground is smooth, θ= 37 °, now a horizontal thrust f is applied to the inclined plane. What should f be to make the object stationary relative to the inclined plane?

The component force of the object under gravity along the inclined plane is mgsin37 = 0.6mg, and the maximum static friction is equal to the sliding friction. F = umgcos37 = 0.2 * 0.8mg = 1.6mg, the object should slide upward along the inclined plane. When the inclined plane is stationary, the object should slide downward along the inclined plane
To keep the object from sliding down,
When the object and the inclined plane have the same minimum acceleration, the object tends to slide downward:
According to Newton's second law: FN sin37 °- μ mgcos37°=ma1； FNcos37°+ μ mgsin37°=mg
Solve the above two equations to obtain A1 = 5m / S ^ 2
Then take M and m as the research object: F1 = (M + m) A1 = 15N
When the object and the inclined body have large acceleration, the object tends to slide upward. Suppose that the downward friction force on the object is the maximum static friction, then FN sin37 ° is obtained from Newton's second law+ μ mgcos37°=ma2； FNcos37°= μ mgsin37°+mg
Solve the above two equations to get A2 = 15.6m/s ^ 2
Then take the object and the whole as the research object: F2 = (M + m) A2 = 46.8n
How large should f be to make the object stationary relative to the inclined plane? 15N

As shown in the figure, the wooden block with mass m = 1kg is stationary on the platform with height h = 1.2m, and the dynamic friction coefficient between the wooden block and the platform is μ= 0.2. Use horizontal thrust f = 20n to make the wood block withdraw when the displacement is S1 = 3M, and the wood block glides out of the platform when S2 = 1m. Calculate the speed of the wood block when it lands. (G is taken as 10m / S2)

Apply the kinetic energy theorem to the whole process of wood block movement:
Fs1- μ mg（s1+s2）+mgh=1
2mv2-0
The solution is: v = 8
2m/s
A: the speed of the wooden block when it falls to the ground is 8
2m/s．