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

Answer

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

$\theta = \omega_o t + \dfrac{1}{2}\alpha t^2$

$\theta =\dfrac{1}{2}\times 11 \times 5^2$

  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

 $a_r = \omega^2 r$

 $a_r = 55^2\times 0.1$

 $a_r =302.5 \ m/s^2$

tangential acceleration of the point on the rim

  $a_t = \alpha r$

  $a_t = 11\times 0.1$

  $a_t = 1.1\ m/s^2$

now, acceleration of the point

$a = \sqrt{a_r^2+a_t^2}$

$a = \sqrt{302.5^2+1.1^2}$

$a = 302.5\ m/s^2$