Figure 10-42 shows a uniform disk that can rotate around its center like a merry-go-round. The disk has a radius of 2.00 cm and a mass of 20.0 grams and is initially at rest. Starting at time t = 0, two forces are to be applied tangentially to the rim as indicated, so that at time t = 1.25 s the disk has an angular velocity of 270 rad/s counterclockwise. Force F1 has a magnitude of 0.160 N. What magnitude is required of force F2?

Yes to the response from 2015, but replace (F1+F2) with (F2-F1)

To find the magnitude of force F2, we can use the formula for torque:

τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Since the disk is initially at rest and reaches an angular velocity of 270 rad/s after 1.25 s, we can find the angular acceleration using the formula:

α = (ωf - ωi) / t

Where ωf is the final angular velocity, ωi is the initial angular velocity (which is 0 in this case), and t is the time taken.

Plugging in the values:

α = (270 rad/s - 0 rad/s) / 1.25 s
α = 216 rad/s²

The moment of inertia for a uniform disk rotating around its center can be calculated using the formula:

I = (1/2) * m * r²

Where m is the mass of the disk and r is the radius.

Plugging in the values:

I = (1/2) * 0.020 kg * (0.02 m)²
I = 0.00002 kg·m²

Now, we can calculate the torque:

τ = Iα
τ = 0.00002 kg·m² * 216 rad/s²
τ = 0.00432 N·m

The torque can also be calculated using the formula:

τ = rF2

Where r is the radius of the disk and F2 is the force applied tangentially to the rim.

Plugging in the values:

0.00432 N·m = (0.02 m)F2
F2 = 0.216 N

Therefore, the magnitude required for force F2 is 0.216 N.

To find the magnitude of force F2, we need to make use of the principle of conservation of angular momentum. The angular momentum of an object is defined as the product of its moment of inertia and its angular velocity.

The moment of inertia (I) of a uniform disk rotating around its center can be calculated using the formula:

I = (1/2) * m * r^2

where m is the mass of the disk and r is its radius.

In this case, the moment of inertia (I) would be:

I = (1/2) * 0.020 kg * (0.02 m)^2 [Converting mass from grams to kilograms]

Now, using the principle of conservation of angular momentum, we can set up the following equation:

I1 * ω1 = I2 * ω2

where I1 and ω1 are the initial moment of inertia and angular velocity of the disk, and I2 and ω2 are the final moment of inertia and angular velocity of the disk.

Substituting the known values:

I1 * ω1 = I2 * ω2

[(1/2) * 0.020 kg * (0.02 m)^2] * 0 = [(1/2) * 0.020 kg * (0.02 m)^2] * 270 rad/s

Simplifying the equation:

0 = (1/2) * 0.020 kg * (0.02 m)^2 * 270 rad/s

Now, we can solve for the magnitude of force F2 using the equation:

F2 * r = ΔL / Δt

where ΔL is the change in angular momentum and Δt is the change in time.

ΔL = I2 * ω2 - I1 * ω1

Δt = 1.25 s - 0 s

Substituting the values:

F2 * 0.02 m = [(1/2) * 0.020 kg * (0.02 m)^2 * 270 rad/s] - [(1/2) * 0.020 kg * (0.02 m)^2 * 0 rad/s]

Simplifying the equation further:

F2 * 0.02 m = (1/2) * 0.020 kg * (0.02 m)^2 * 270 rad/s

Now, we can solve for the magnitude of force F2 by dividing both sides of the equation by 0.02 m:

F2 = [(1/2) * 0.020 kg * (0.02 m)^2 * 270 rad/s] / 0.02 m

F2 = 0.020 kg * (0.02 m) * 270 rad/s

Calculating the expression:

F2 = 0.020 kg * 0.02 m * 270 rad/s

F2 = 0.108 N

Therefore, the magnitude of force F2 required is 0.108 N.

Use Summation of

Torque(T) = Inertia(I)*Acceleration(alpha)

T=I*alpha, and we know I = (1/2)(m)(r)^2
T=(F1+F2)r
alpha= dw/dt= 270rad/1.25s ( w is the given angular velocity)
so (F1+F2)*r=(1/2)(m)(r)^2 * alpha
solve for F2...