A 175 kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s?
I first solved for the angular acceleration and got 0.2 rev/s^2
Then I used 1/2mr*(angular acceleration)
and got 26.25 N, but that answers is incorrect.
Please help, thanks!
Let "omega" = w = angular velocity
1 rev/s = 2pi rad/s
angular velocity = w = (0.400)(2pi rad/s) = 2.5133 rad/s
Linear velocity = v = wr
v = (2.5133 rad/s)(1.5m/rad) = 3.77 m/s
Linear accel. = (change in velocity)/time = a
a = (Vf - Vi)/t = (3.77 m/s - 0)/2.00 s = ______?
Force = mass x acceleration = ____?
(use mass given and the value of a)
neither one of these are correct
To find the constant force required to bring the merry-go-round from rest to the given angular speed, we can use the following steps:
Step 1: Convert the given angular speed from revolutions per second (rev/s) to radians per second (rad/s). Since 1 revolution is equal to 2π radians, we have:
Angular speed = 0.400 rev/s * (2π rad/1 rev) = 0.400 * 2π rad/s
Step 2: Use the formula for angular acceleration, which is the change in angular velocity divided by the change in time:
Angular acceleration (α) = (angular speed - initial angular speed) / time
In this case, the initial angular speed is 0 since the merry-go-round is at rest. Thus:
α = (0.400 * 2π rad/s - 0 rad/s) / 2.00 s
Step 3: Compute the torque τ (the product of force and lever arm) acting on the merry-go-round using the formula:
τ = moment of inertia (I) * angular acceleration (α)
The moment of inertia of a solid disk can be calculated as I = (1/2) * m * r^2, where m is the mass of the disk and r is its radius.
I = (1/2) * 175 kg * (1.50 m)^2
Step 4: Finally, find the force required by rearranging the torque formula:
τ = F * r, where F is the force applied at a radius r.
Therefore, F = τ / r
Now, let's plug in the values and calculate the force:
I = (1/2) * 175 kg * (1.50 m)^2 = 196.875 kg∙m^2
α = (0.400 * 2π rad/s - 0 rad/s) / 2.00 s = 1.2608 rad/s^2
τ = 196.875 kg∙m^2 * 1.2608 rad/s^2 = 248.321 N∙m
r = 1.50 m (given)
F = 248.321 N∙m / 1.50 m ≈ 165.54 N
Therefore, the constant force required to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s is approximately 165.54 Newtons.