Question

The rate of rotation of a solid disk with a radius of `6 m` and mass of `5 kg` constantly changes from `2 Hz` to `12 Hz`. If the change in rotational frequency occurs over `5 s`, what torque was applied to the disk?

Answer 1

`tau~~1131"Nm"` (counterclockwise)

Explanation:

Torque can be expressed by the following equation:

`color(skyblue)(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"=5"kg"`
• `|->"R"=6"m"`
• `|->"f"_i=2"s"^-1`
• `|->"f"_f=12"s"^-1`
• `|->Deltat=5"s"`

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

`tau=1/2(5"kg")(6"m")^2*(2pi(12"s"^-1-2"s"^-1)/(5"s"))`

`=1130.973"Nm"`

`~~1131"Nm"`

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