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

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 m s - 2 . (Take g = 10 m s - 2 ) If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

TestHub · Accepted Answer

The forces acting on the solid sphere is shown below.
Here, $N$ is the normal force acting on sphere and weight $m g$ is acting downwards.

Taking torque about contact point.
$\vec{τ} = m g R \sin 30^{\otimes} + 1 \times 1^{⊙}$
$= 10 \times 1 \times \frac{1}{2} - 1$ (Taking $\otimes$ as positive)
Then, we have
$\Rightarrow 5 - 1 = I_{sphere about tangent} α$
Using parallel axis theorem,
$τ = (\frac{2}{5} m R^{2} + m R^{2}) α = \frac{7}{5} m R^{2} α$
$\Rightarrow α = \frac{20}{7} rad s^{- 2}$
So, acceleration of sphere down the plane is $a_{cm} = α R = \frac{20}{7} m s^{- 2}$
$a_{cm} = 2.86 m s^{- 2}$

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