Question

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating about a fixed axis. The applied force acts for 5.80 s. During this time, the angular speed of the wheel increases from 0 to 10.4 rad/s. The applied force is then removed, and the wheel comes to rest in 60.4 s. (a) Find the moment of inertia of the wheel. kg · m2 (b) Find the magnitude of the torque due to friction. N · m (c) Find the total number of revolutions of the wheel during the entire interval of 66.2 s.

Answer 1

a)
`I = 19.799\,kg\cdot m^(2)`, b)
`T = -3.405\,N\cdot m`, c)
`n_(T) \approx 54.842\,rev`

Step-by-step explanation:

a) The net torque is:

`T = I\cdot \alpha`

Let assume a constant angular acceleration, which is:

`\alpha = (\omega-\omega_(o))/(t)`

`\alpha = (10.4\,(rad)/(s) - 0\,(rad)/(s) )/(5.80\,s)`

`\alpha = 1.793\,(rad)/(s^(2))`

The moment of inertia of the wheel is:

`I = (T)/(\alpha)`

`I = (35.5\,N\cdot m)/(1.793\,(rad)/(s^(2)) )`

`I = 19.799\,kg\cdot m^(2)`

b) The deceleration of the wheel is due to the friction force. The deceleration is:

`\alpha = (\omega-\omega_(o))/(t)`

`\alpha = (0\,(rad)/(s) - 10.4\,(rad)/(s))/(60.4\,s)`

`\alpha = - 0.172\,(rad)/(s^(2))`

The magnitude of the torque due to friction:

`T = (19.799\,kg\cdot m^(2))\cdot (-0.172\,(rad)/(s^(2)) )`

`T = -3.405\,N\cdot m`

c) The total angular displacement is:

`\theta_(T) = \theta_(1) + \theta_(2)`

`\theta_(T) = ((10.4\,(rad)/(s) )^(2)-(0\,(rad)/(s) )^(2))/(2\cdot (1.793\,(rad)/(s^(2)) )) + ((0\,(rad)/(s) )^(2)-(10.4\,(rad)/(s) )^(2))/(2\cdot (-0.172\,(rad)/(s^(2)) ))`

`\theta_(T) = 344.580\,rad`

The total number of revolutions of the wheel is:

`n_(T) = (\theta_(T))/(2\pi)`

`n_(T) = (344.580\,rad)/(2\pi)`

`n_(T) \approx 54.842\,rev`