The combination of an applied force and a friction force produces a constant total torque of 35.9 n·m on a wheel rotating about a fixed axis. the applied force acts for 6.20 s. during this time, the angular speed of the wheel increases from 0 to 9.9 rad/s. the applied force is then removed, and the wheel comes to rest in 60.7 s. a. find the moment of inertia of the wheel. b. find the magnitude of the torque due to friction c. find the total number of revolutions of the wheel during the entire interval of 66.9 s.

Explanation:

Given that,

Torque = 35.9 N-m

Time of acceleration= 6.20 s

Initial angular speed = 0 rad/s

Final angular speed = 9.9 rad/s

Time of declaration = 60.7 s

(a). We need to calculate the angular acceleration

Using formula of angular acceleration

Put the value into the formula

We need to calculate the moment of inertia of the wheel

Using formula of moment of inertia

Put the value into the formula

The moment of inertia of the wheel is

(b). The applied force is removed, then the magnitude of the torque due to friction

We need to the magnitude of the torque due to friction

Using equation of rotation

Put the value into the formula

Magnitude of angular acceleration is 0.163 rad/s²

We need to calculate the frictional torque

Using formula of torque

Put the value into the formula

The frictional torque is

(c). We need to calculate  the total number of revolutions of the wheel during the entire interval of 66.9 s

Using formula of angular displacement

During the applied force,

Put the value into the formula

During the removed of force,

Put the value into the formula

The total angular displacement

The total number of revolutions of the wheel is 52.67 rev.

Hence, This is the required solution.

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).