A solid sphere of mass 1kg and radius 1m rolls without slipping on...

The forces acting on the solid sphere is shown below.

Here, $N$ is the normal force acting on sphere and weight $mg$ is acting downwards.

Taking torque about contact point.

$\overset{\to}{τ} = mgRsin{30}^{⊗} + 1 \times {1}^{⊙}$

$= 10 \times 1 \times \dfrac{1}{2} - 1$ (Taking $⊗$ as positive)

Then, we have

$⇒ 5 - 1 = {I}_{\text{sphere about tangent}}\alpha$

Using parallel axis theorem,

$τ = \left(\dfrac{2}{5}m{R}^{2} + m{R}^{2}\right)\alpha = \dfrac{7}{5}m{R}^{2}\alpha$

$⇒ \alpha = \dfrac{20}{7}rad{s}^{- 2}$

So, acceleration of sphere down the plane is ${a}_{cm} = \alphaR = \dfrac{20}{7}m{s}^{- 2}$

${a}_{cm} = 2 . 86m{s}^{- 2}$