Question

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed //tex.z-dn.net/?f=v_%7Bcom%7D" id="TexFormula1" title="v_{com}" alt="v_{com}" align="absmiddle" class="latex-formula">,0 = 9.0 m/s and initial angular speed ω₀ = 0. The coefficient of kinetic friction between the ball and the lane is 0.23. The kinetic frictional force f with $\bar{k}$ acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed $v_{com}$ has decreased enough and angular speed ω has increased enough, the ball stops sliding and then rolls smoothly.
(a) What then is $v_{cm}$ in terms of ω? m·ω
(b) During the sliding, what is the ball's linear acceleration? m/s²
(c) During the sliding, what is the ball's angular acceleration? rad/s²
(d) How long does the ball slide? s
(e) How far does the ball slide? m
(f) What is the speed of the ball when smooth rolling begins? m/s

Answer 1

Answer:

a) v_com= Rω

b) -2.254 m/$s^{2}$

c) 51.2 rad/$s^{2}$

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

$\alpha$=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b)    F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

c) α=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec