A bowling ball rolis without slipping down an inclined plane inclined at an angle θ to the horizontal, as shown. The coefficient of static friction between the ball and the surface is μs, and the coefficient of kinetic friction is μk. What is the magnitude of the force of friction acting on the ball?

Question

A bowling ball rolis without slipping down an inclined plane inclined at an angle θ to the horizontal, as shown. The coefficient of static friction between the ball and the surface is μs, and the coefficient of kinetic friction is μk. What is the magnitude of the force of friction acting on the ball? (A) zero (B) μs m g cos θ (C) μs (7/5) m g sin θ (D) (2/7) m g sin θ

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1776f35e8a4f42ceb24383ee38419997-.png" /> m g sin θ-f=m a cm f R=(2/5) M R2 α (torque =I α about COM ) a cm =R α ⇒ f=(2 m g sin θ/7)

zero

$μ_{s} m g cos θ$

$μ_{s} \frac{7}{5} m g sin θ$

$\frac{2}{7} m g sin θ$

$m g sin θ - f = m a_{c m}$
$f R = \frac{2}{5} M R^{2} α$
(torque $= I α$ about COM $)$
$a_{c m} = R α \Rightarrow f = \frac{2 m g s i n θ}{7}$

zero

μs​mgcosθ

μs​57​mgsinθ

72​mgsinθ

mgsinθ−f=macm​ fR=52​MR2α (torque =Iα about COM ) acm​=Rα⇒f=72mgsinθ​

$μ_{s} m g cos θ$

$μ_{s} \frac{7}{5} m g sin θ$

$\frac{2}{7} m g sin θ$

$m g sin θ - f = m a_{c m}$
$f R = \frac{2}{5} M R^{2} α$
(torque $= I α$ about COM $)$
$a_{c m} = R α \Rightarrow f = \frac{2 m g s i n θ}{7}$