Question

The rate of rotation of a solid disk with a radius of `8 m` and mass of `2 kg` constantly changes from `2 Hz` to `8 Hz`. If the change in rotational frequency occurs over `12 s`, what torque was applied to the disk?

Answer 1

`tau=201"Nm"` (counterclockwise)

Explanation:

Torque can be expressed by the following equation:

`color(darkblue)(tau=Ialpha)`

where `I` is the moment of inertia of the object and `alpha` is its angular acceleration

Since a specific `I` is not provided for the disk, it will be assumed that it is a uniform disk which rotates about a center axis. Therefore, we can assume `color(darkblue)(I=1/2MR^2)`.

The average angular acceleration is given by:

`color(red)(alpha_(avg)=(Deltaomega)/(Deltat))`

where `omega` is the angular velocity of the disk and `Deltat` is the period over which the change occurs

Since `omega=2pif`, we can write the angular acceleration in terms of the frequency as:

`color(red)(alpha_(avg)=(2pi(f_f-f_i))/(Deltat))`

Putting all of the above together, we have the equation:

`color(darkblue)(tau=1/2MR^2*(2pi(f_f-f_i))/(Deltat))`

We are given the following information:

• `|->"M"=2"kg"`
• `|->"R"=8"m"`
• `|->"f"_i=2"s"^-1`
• `|->"f"_f=8"s"^-1`
• `|->Deltat=12"s"`

Substituting these values into the equation we derived above, we have:

`tau=1/2(2"kg")(8"m")^2*(2pi(8"s"^-1-2"s"^-1))/(12"s")`

`=201.062"Nm"`

`~~201"Nm"`

Note this occurs in the counterclockwise direction as the convention is that counterclockwise is positive.