A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ,ms -1 goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ?

Question

A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ,ms -1 goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/facef1e7f47f6266bd9aa3cadd854a9e-.png" /> (A) 0.60 s (B) 0.52 s (C) 0.57 s (D) 0.80 s

Tardigrade Admin · Accepted Answer

a =( g sin θ/1+( I ) mR 2)=(5/7) × (10/2)=(25/7) =(2 v0/a)=(2 × 1 × 7/25) =0.56

Q. A sphere of mass $2 k g$ and radius $0.5 m$ is rolling with an initial speed of $1, m s^{- 1}$ goes up an inclined plane which makes an angle of $3 0^{\circ}$ with the horizontal plane, without slipping. How low will the sphere take to return to the starting point $A$ ?

0.60 s

0.52 s

0.57 s

0.80 s

$a = \frac{g s i n θ}{1 + \frac{I}{m R ^{2}}} = \frac{5}{7} \times \frac{10}{2} = \frac{25}{7}$
$= \frac{2 v _{0}}{a} = \frac{2 \times 1 \times 7}{25}$
$= 0.56$

Q. A sphere of mass 2kg and radius 0.5m is rolling with an initial speed of 1 ,ms−1 goes up an inclined plane which makes an angle of 30∘ with the horizontal plane, without slipping. How low will the sphere take to return to the starting point A ?