Question

A motor produces a torque of 22 Nm. It is used to accelerate a wheel of radius 10cm and moment of inertia 2 kg.m2 which is initially at rest. Calculate

a)

the number of revolutions made by the wheel in the first 5s

b)

the angular velocity after 5s

c)

the acceleration of a point on the rim of the wheel after 5s

Answer 1

given,

torque produced, τ = 22 N.m

Radius of the wheel. r = 10 cm

Moment of inertial = 2 kg.m²

initial angular speed = 0 rad/s

time, t = 5 s

a) we know,

τ = I α

22 = 2 x α

α = 11 rad/s²

using equation of rotation motion

`\theta = \omega_o t + (1)/(2)\alpha t^2`

`\theta =(1)/(2)* 11 * 5^2`

θ = 137.5 rad

θ = 137.5/2π = 22 revolution.

b) angular velocity of the motor

`\omega_f = \omega_i + \alpha t`

`\omega_f = 0 + 11 x 5`

`\omega_f = 55\ rad/s`

c) acceleration of a point on the rim of the wheel

radial acceleration

`a_r = \omega^2 r`

`a_r = 55^2* 0.1`

`a_r =302.5 \ m/s^2`

tangential acceleration of the point on the rim

`a_t = \alpha r`

`a_t = 11* 0.1`

`a_t = 1.1\ m/s^2`

now, acceleration of the point

`a = √(a_r^2+a_t^2)`

`a = √(302.5^2+1.1^2)`

`a = 302.5\ m/s^2`