__Question 7.14__:

__Solution__:*m*= 3 kg

Radius of the hollow cylinder,

*r*= 40 cm = 0.4 m

Applied force,

*F*= 30 N

The moment of inertia of the hollow cylinder about its geometric axis:

*I*=

*mr*

^{2}

= 3 × (0.4)

^{2}= 0.48 kg m

^{2}

Torque, τ = F X r = 30 X 0.4 = 12 Nm

For angular acceleration α, torque is also given by the relation:

τ = Iα

α = τ / I = 12 / 0.48 = 25 rad s

^{-2}

Linear acceleration = τα = 0.4 × 25 = 10 m s

^{–2}

## No comments:

## Post a Comment