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At t=0 the disk is also spinning counter-clockwise with angular velocity w_o. Thus, when viewing the disk at time t=0 you will see it move to the right (I call this the + x-direction) but it is spinning counter-clockwise.

Question A

There comes a time, t, that the disk will come to a halt. How far will the disk have moved (since t=0) when it comes to a halt?

Question B

What is the angular velocity, w, of the disk at time t when it comes to a halt. If the angular velocity is clockwise indicate that with a + sign, if it is counter-clockwise, indicate that with a - sign. If it is zero, then answer w=0

question C

There comes a time (t_1) that the angular velocity will become zero, thus the disk will not rotate anymore. How far will the disk have traveled in the -x direction when t=t_1?

question D

What will be the velocity of the center of mass at t_1? Signs in the velocity are important unless the speed is zero.

question E

If we compare the w was -w_o and when the velocity of the center of mass was +w_o*R/4, how much kinetic energy has the disk lost at time t_1?

i have calculated first two parts but couldn't figure out the rest three

So the first two and the equations I have considered are as follows

initial conditions

solid disk has radius R and mass M

at time t=0

omega is -w_o (counter clock wise direction)

v_o=+w_o*R/4 translational motion of the center of mass in the + direction

Mg*b is the friction between disk and the surface. The disk is rotating

counter-clockwise AND the center of mass is moving in the + direction.

Thus the frictional force is in the - direction.

v_t=v_o+at

a=-Mg*b/M=-g*b (a is in the - direction)

ASNSWER TO QUESTION A

When the object comes to a halt, all translational KE has been converted

into heat. The frictional force Mg*b is constant throughout the sliding.

Thus the KE of translation [0.5*M(w_o*R)^2]/16=Mg*b*x

Thus x (distance traveled) is(w_o*R)^2]/32g*b

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QUESTION B

The disk will move in the + direction but it will stand still when

0=w_o*R/4 - g*b*t

thus the disk will come to a halt at t=w_o*R/4g*b (eq 1)

If at that time the angular velocity (w) is still negative, the object

will roll back in the negative direction where it came from. If omega

is positive the disk will continue to move in the + direction

(clockwise rotation).

w_t= -w_o + alpha*t (eq 2)

alpha is dw/dt

M*g*b*R=I*alpha (torque equation)

I_disk=(M*R^2)/2

alpha=gb*2R)/(R^2)= g*b*2/R (eq 3)

Substitute t of (eq 1) into (eq 2)

Substitute alpha (eq 3) in (eq 2)

ANSWER TO B

You then find

w_t= -w_o + w_o/2= -w_o/2

Conclusion the disk is now standing still and it is rotating at half

the angular velocity than at time t=0 but it is still rotating counter clockwise as it will roll back

Please I need some hints to solve for other parts