Bowling Ball Slipping and Rolling: Analyzing Acceleration

[jnimagine]

A uniform, spherical bowling ball of mass m and radius R is projected horizontally along the
floor at an initial velocity v0 = 6.00 m/s. The ball is not rotating initially, so w0 = 0. It
picks up rotation due to (kinetic) friction as it initially slips along the floor. The coefficient of
kinetic friction between the ball and the floor is μk. After a time ts, the ball stops slipping and makes a transition to rolling without slipping at angular speed ws and translational velocity _s. Thereafter, it rolls without slipping at constant velocity.

(b) Find an equation for the linear acceleration a of the ball during this time. The acceleration should be negative, since the ball is slowing down.
(c) Find an equation for the angular acceleration a of the ball while it is slipping. It will be
simpler if you use the sign convention that clockwise rotations are positive, so  > 0.
(d) What constraint on w and v must take effect at time t = ts, the moment when the ball
stops slipping and begins rolling without slipping?

Here is my attempt:
b) slipping = rw + deltavt = vt
and then you get a derivative of it to get a = u_kg
c) a = torque / I
r(ru_kmg / 2/5mr^2) + dv/dt = -u_kg
and we get like -7/2u_kg from this...
d)...

please help!

 

[Doc Al]

Friction is the only horizontal force acting on the ball. Use Newton's 2nd law for translation and rotation to find the linear and angular accelerations.

For (d), what's the condition for rolling without slipping?

 

[thomate1]

Hello,

Thank you for sharing your attempt at solving this problem. Here is a possible response to the questions:

b) To find the linear acceleration, we can use the equation: a = (vf - vi)/t. Since the ball is slowing down, vf = 0 m/s and vi = 6.00 m/s. We can also express the final time as ts, so the equation becomes: a = (0 - 6.00)/ts. This simplifies to a = -6.00/ts.

c) To find the angular acceleration while the ball is slipping, we can use the equation: a = τ/I. The torque acting on the ball is due to the friction force, which is given by μkmg. The moment of inertia for a uniform spherical ball is 2/5mr^2. Therefore, the equation becomes: a = (μkmg)/((2/5)mr^2). Using the sign convention that clockwise rotations are positive, we can express this as a = -μk(5g)/(2r).

d) At time t = ts, the ball stops slipping and starts rolling without slipping. This means that the linear velocity and angular velocity must be equal at this point. Mathematically, this can be expressed as: v = ωr. Since the ball is rolling without slipping after this point, we can also say that the linear velocity is equal to the product of the angular velocity and the radius: v = ωr. Combining these two equations, we get: ωr = ωr. This constraint ensures that the ball rolls without slipping.

I hope this helps. Keep up the good work!