Question

A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle \(\theta\) .The frictional force 

• A. converts translational energy to rotational energy
• B. dissipates energy as heat
• C. decreases the rotational motion
• D. decreases the rotational and translational motion

Answer 1

The Correct Option is A

We have,
mg\(\times\)sinA−f=ma ________________________ (1)
in pure rolling \(\alpha\) =\(\frac{a}{r}\)
\(\alpha\) = angular acceleration 
Now,τ=I\(\alpha\)  
I = moment of inertia 
f\(\times\)r = I\(\alpha\)
f\(\times\)r = I\(\frac{a}{r}\) 
\(\therefore\) f =Iar² ________________________________ (2)
Substituting (2)in (1)

we have mgsinA−I\(\frac{a}{r^2}\)=ma

here I of solid cylinder=\(\frac{mr^2}{2}\) 

mgsinA−\(\frac{ma}{2}\)=ma 

a=\(\frac{(2gsinA)}{3}\)
Now 
from (2) we have 
f =\(\frac{la}{r^2}\)
f =\(\frac{ma}{2}\) which should be less than or equal to μN otherwise body will slip (where u is coefficient of friction and N is normal acting on cylinder which is equal to mg cos A) 

\(\frac{ma}{2}\)\(\leq\) \(\mu\)mgcosA

putting value of a

\(\frac{(m\times g \times \sin A)}{3}\) \(\leq\) \(\mu\)mgcosA 
tanA  \(\leq\) 3μ

Therefore, the correct option is (A): converts translational energy to rotational energy