You wish to accelerate a small merry-go-round from rest to a rotational speed of one-fourth of a revolution per second by pushing tangentially on it. Assume the merry-go-round is a disk with a mass of 150 kg and a radius of 1.80 m. Ignoring friction, how hard do you have to push tangentially to accomplish this in 8.00 s? (Use energy methods and assume a constant push on your part.)
To calculate the force required to accelerate the merry-go-round, we can use the principles of energy conservation. The kinetic energy of the rotating disk is given by:
KE = (1/2) I ω^2,
where KE is the kinetic energy, I is the moment of inertia of the disk, and ω is the angular velocity.
The moment of inertia of a disk is given by:
I = (1/2) m r^2,
where m is the mass of the disk and r is its radius.
Given that the mass (m) of the merry-go-round is 150 kg and the radius (r) is 1.80 m, we can calculate the moment of inertia (I):
I = (1/2) × 150 kg × (1.80 m)^2 = 243 kg·m^2.
The desired angular velocity (ω) is one-fourth of a revolution per second, which can be converted to radians per second:
ω = (1/4) revolution per second × 2π radians per revolution = π/2 radians per second.
Now we can calculate the initial kinetic energy (KE_initial) and the final kinetic energy (KE_final) of the disk:
KE_initial = (1/2) × 243 kg·m^2 × 0^2 = 0.
KE_final = (1/2) × 243 kg·m^2 × (π/2)^2 = 74.07 joules.
Since energy is conserved, the work done by the push will be equal to the change in kinetic energy:
Work = KE_final - KE_initial.
To find the force required, we need to divide the work done by the push by the distance over which it is applied:
Work = Force × Distance.
In this case, the distance is equal to the circumference of the merry-go-round, which is 2πr:
Force × 2πr = KE_final - KE_initial.
Finally, we can solve for the force:
Force = (KE_final - KE_initial) / (2πr).
Substituting the known values:
Force = (74.07 joules - 0 joules) / (2π × 1.80 m).
Evaluating this expression gives the force required to be approximately 6.48 newtons.