A solid disc with radius R is is spinning. An object exerts a force F at the edge of the disc to slow down the disc. The disc slows down to rest in 2.0 seconds. How long would the disc take to stop if:

the mass of the disc doubled in s
the radius of the disc doubled in s
the force doubled in s

F=ma, so:

the mass of the disc doubled in s - it would take twice as long to stop

the radius of the disc doubled in s - it would take twice as long to stop

the force doubled in s - it would take half the time to stop

To determine how each of these changes affects the time it takes for the disc to stop, we need to understand the relationship between force, mass, radius, and rotational motion.

The rotational motion of the disc is described by the equation:

torque = moment of inertia * angular acceleration

The moment of inertia (I) depends on the mass (m) and the radius (R) of the disc:

moment of inertia (I) = 1/2 * m * R^2

The angular acceleration (α) is the rate at which the disc slows down, and it is related to the net torque applied to the disc (τ) by the equation:

τ = I * α

The net torque exerted on the disc is equal to the force (F) applied at the edge of the disc multiplied by the radius (R):

τ = F * R

We can rearrange these equations to find the angular acceleration and solve for the time it takes for the disc to stop.

For the given time of 2.0 seconds, let's calculate the original values:

moment of inertia (I_0) = 1/2 * m * R^2
angular acceleration (α_0) = τ / I_0 = F * R / (1/2 * m * R^2) = 2F/m
time to stop (t_0) = 2 seconds

Case 1: The mass of the disc doubles:

In this case, the new moment of inertia (I_1) will be:

I_1 = 1/2 * 2m * R^2 = m * R^2

The new angular acceleration (α_1) will be:

α_1 = τ / I_1 = F * R / (m * R^2) = F / (m * R)

The time it takes for the disc to stop in this case (t_1) can be calculated using the new angular acceleration:

t_1 = α_0 / α_1 = (2F/m) / (F / (m * R)) = 2R

Therefore, if the mass of the disc doubles, it will take twice as long for the disc to stop. So, the new time to stop, t_1, will be 4.0 seconds.

Case 2: The radius of the disc doubles:

In this case, the new moment of inertia (I_2) will be:

I_2 = 1/2 * m * (2R)^2 = 2m * R^2

The new angular acceleration (α_2) will be:

α_2 = τ / I_2 = F * R / (2m * R^2) = F / (2m)

The time it takes for the disc to stop in this case (t_2) can be calculated using the new angular acceleration:

t_2 = α_0 / α_2 = (2F/m) / (F / (2m)) = 1/4

Therefore, if the radius of the disc doubles, it will take one-fourth of the original time for the disc to stop. So, the new time to stop, t_2, will be 0.5 seconds.

Case 3: The force doubles:

In this case, the new angular acceleration (α_3) will be:

α_3 = τ / I_0 = (2F) * R / (1/2 * m * R^2) = 4F/m

The time it takes for the disc to stop in this case (t_3) can be calculated using the new angular acceleration:

t_3 = α_0 / α_3 = (2F/m) / (4F/m) = 1/2

Therefore, if the force doubles, it will take half of the original time for the disc to stop. So, the new time to stop, t_3, will be 1.0 second.

To summarize:

- If the mass of the disc doubles, it will take 4.0 seconds for the disc to stop.
- If the radius of the disc doubles, it will take 0.5 seconds for the disc to stop.
- If the force doubles, it will take 1.0 second for the disc to stop.