A bowling ball of mass M and radius R is thrown along a level surface so that initially ( t= 0) it slides with a linear speed but does not rotate, see the figure. As it slides, it begins to spin, and eventually rolls without slipping.
How far has the ball moved down the lane when it starts rolling without slipping?
What are its final linear speed?
What are its final rotational speed?
Thanks:]
4 answers
You need to specify kinetic friction coefficient and the initial velocity.
it says to answer it in terms of those coefficients
like, it's asking for formulas...
Initial w, angular velocity, is zero.
now while sliding
torque = I alpha
mu m g r = (2/5) m r^2 alpha
alpha = (5/2) mu g / r
and
F = m a
mu m g = m a
a = mu g
w = alpha t
v = Vi - a t
when does w r equal v (no slip)
r alpha t = v = Vi - a t
(5/2) mu g t = Vi - mu g t
(7/2)mu g t = Vi
so
t = (2/7)Vi/(mu g)
that is the time sliding
work back up to get distance and final v
Check that with energy argument:
frictional work done = mu m g d
Initial Ke - frictional work = Final Ke
(1/2) m Vi^2 - mu m g d = (1/2) m v^2 +(1/2) (2/5) m r^2 (v^2/r^2)
Vi^2 - 2 mu g d = (7/5)v^2
now while sliding
torque = I alpha
mu m g r = (2/5) m r^2 alpha
alpha = (5/2) mu g / r
and
F = m a
mu m g = m a
a = mu g
w = alpha t
v = Vi - a t
when does w r equal v (no slip)
r alpha t = v = Vi - a t
(5/2) mu g t = Vi - mu g t
(7/2)mu g t = Vi
so
t = (2/7)Vi/(mu g)
that is the time sliding
work back up to get distance and final v
Check that with energy argument:
frictional work done = mu m g d
Initial Ke - frictional work = Final Ke
(1/2) m Vi^2 - mu m g d = (1/2) m v^2 +(1/2) (2/5) m r^2 (v^2/r^2)
Vi^2 - 2 mu g d = (7/5)v^2
2 answers