Question

You throw a bowling ball of mass M and radius 1m has an initial speed v0 down the lane. It has an initial angular velocity of 3v0/R just after its release. The coefficient of kinetic friction is 0.08. It begins to roll without slipping at t=2s. What is the speed of the ball just as it begins rolling without slipping?

Answer 1

Answer:

Explanation:

Given

Initial angular velocity $\omega_0=\frac{3v_0}{R}$

Initial speed $u=v_0$

Coefficient of kinetic friction $\mu _k=0.08$

As No external torque is applied therefore angular momentum is conserved

$L_i=L_f$

$I_{cm}v_i+Mv_{cm}_iR=I_{cm}\omega _f+Mv_{cm}_f\cdot R$

For rolling without slipping $v_f=\omega _fR$

$\frac{2}{5}MR^2\cdot (\frac{3v_0}{R})+Mv_0\cdot R=\frac{2}{5}MR^2\cdot \frac{v_f}{R}+Mv_f\cdot R$

$\frac{6}{5}Mv_0R+Mv_0R=\frac{2}{5}Mv_fR+Mv_f\cdot R$

$v_f=\frac{11}{7}v_0$

acceleration provided by surface

$a=\mu _kg$

$a=0.08\times 9.8$

$a=0.784\ m/s^2$

Speed of ball after $t=2\ s$

using $v=u+at$

v=final velocity

u=initial velocity

$\frac{11}{7}v_0=v_0+0.784\times 2$

$\frac{11}{7}v_0-v_0=0.784\times 2$

$v_0=2.744\ m/s$

thus $v_f=\frac{11}{7}\times 2.74$

$v_f=4.312\ m/s$