Question

A motor drives a disk initially at rest through 25.9 rotations in 5.0 s. Assume the vector sum of the torques caused by the force exerted by the motor and the force of friction is constant. The rotational inertia of the disk is 4.0 kg⋅m2. When the motor is switched off, the disk comes to rest in 12 s.
A.)What is the magnitude of torque created by the force of friction?
B.)What is the magnitude of torque caused by the force exerted by the motor?

Answer 1

Answer

given,

rotation of disk is equal to(θ) = 25.9

time taken = 5 s

moment of inertia of disk = 4 Kg.m²

disk come to rest in = 12 sec

a)  θ = 2 π x 25.9

    θ = 162.73 rad

    using equation of rotational motion

     θ =  ω₀t + 0.5 α t²

     162.79 =  0 + 0.5 x α x 5²

      α = 13.02 rad/s²

total torque acting

 $\tau_{frictional} + \tau_{rotational} = I \alpha$

 $\tau_{frictional} + \tau_{rotational} = 4\times 13.02$

 $\tau_{frictional} + \tau_{rotational} =52.08$

again using the rotational equation

   $\omega_f - \omega_i = \alpha\ t$

   $\omega_f -0 = 13.02 \times 5$

   $\omega_f = 65.1\ rad/s$

when only friction is acting angular acceleration

   $\omega_f - \omega_i = \alpha\ t$

   $0 -65.1 =\alpha\ 12$

   $\alpha= -5.425\ rad/s^2$    

now torque due to friction

  $\tau_{frictional} =- 4 \times 5.425$

  $\tau_{frictional} =-21.7\ Nm$

now,

 $-21.7+ \tau_{rotational} =52.08$

 $\tau_{rotational} =73.78\ Nm$