Question

A cord is wrapped around the rim of a wheel 0.235 m in radius, and a steady pull of 43.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center.
The moment of inertia of the wheel about this shaft is 5.15 kg⋅m2 Compute the angular acceleration of the wheel.

Answer 1

Answer:

$\alpha = 1.96\ rad/s^2$

Explanation:

given,

radius of wheel = 0.235 m

steady pull = 43 N

moment of inertia of wheel = 5.15 Kg.m²

angular acceleration  = ?

torque = Force x distance

τ = F r

and torque is also known as

τ = I α

computing both the torque expression

I α  = F r

$\alpha = \dfrac{Fr}{I}$

$\alpha = \dfrac{43 \times 0.235}{5.15}$

$\alpha = 1.96\ rad/s^2$

hence, the angular acceleration of wheel is equal to $\alpha = 1.96\ rad/s^2$