A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the instantaneous speed of point B with respect to the ground?
A)zero m/s
B)5.0 m/s
C)7.1 m/s
D)7.5 m/s
E)10.0 m/s

Question

A 2.0-kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with speed 5.0 m/s.  
The point A is the uppermost point on the disk and the point B is along the horizontal line that connects the center of the disk to the rim.
-What is the instantaneous speed of point B with respect to the ground?
A)zero m/s 
B)5.0 m/s 
C)7.1 m/s 
D)7.5 m/s 
E)10.0 m/s

Quizplus · Accepted Answer

The instantaneous speed of point B can be found using the equation v = ωr, where ω is the angular velocity and r is the distance from the center of rotation to point B. Since the disk is rolling without slipping, we know that the linear velocity of the center of the disk is v = 5.0 m/s, and the angular velocity is ω = v/r.

To find r, we can use the fact that the center of the disk and point B lie on a horizontal line, so the distance between them is the radius of the disk. The radius of the disk is not provided, but we can calculate it using the moment of inertia of a solid disk about its center, I = (1/2)mr^2.

Solving for r, we have r = sqrt(2I/m) = sqrt(1/2) * sqrt(I/m) = (1/sqrt(2)) * sqrt((1/2) * (2.0 kg) * (0.5 m)^2) = 0.5 m.

Thus, ω = v/r = (5.0 m/s) / (0.5 m) = 10.0 rad/s, and vB = ωr = (10.0 rad/s) * (0.5 m) = 5.0 m/s. Therefore, the instantaneous speed of point B with respect to the ground is 5.0 m/s.

A 20-Kg Solid Disk Rolls Without Slipping on a Horizontal Surface