A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance from the center

Question

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination $θ = 30 °$ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance $r = 0.5 m$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is _______ ${ms}^{- 2}$ . (Take $g = 10 {ms}^{- 2}$ ) [2022]

Smart Achievers · Accepted Answer

(2.86)

We know that for a rolling body, point of contact is instantaneous axis of rotation.

$So, τ_{point of contact} = I α$

$\Rightarrow 1 \times 0.5 + (m g \sin 30 °) R - 1.5 \times 1 = I α$

$\Rightarrow 0.5 + 5 R - 1.5 = I α$

$\Rightarrow 5 R - 1 = \frac{7}{5} m R^{2} α$

$\Rightarrow 4 = \frac{7}{5} α$

$\Rightarrow α = \frac{20}{7} {rad/s}^{2}$

$So, a_{c m} = α R = \frac{20}{7} \times 1 = \frac{20}{7} {m/s}^{2} = 2.86 {m/s}^{2}$

Solution

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Solution

Ans. (2.86) We know that for a rolling body, point of contact is instantaneous axis of rotation. So, τpoint of contact=Iα ⇒1×0.5+(mgsin30°)R-1.5×1=Iα ⇒0.5+5R-1.5=Iα ⇒5R-1=75mR2α ⇒4=75α ⇒α=207 rad/s2 So, acm=αR=207×1=207m/s2=2.86 m/s2

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