Question

A 140-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 0.800 rev/s in 2.00 s

Answer 1

Answer:

The constant force is 263.55 newtons

Explanation:

There's a rotational version of the Newton's second law that relates the net torque on an object with its angular acceleration by the equation:

$\tau = I\alpha$ (1)

with τ the net torque and α the angular acceleration. It’s interesting to note the similarity of that equation with the well-known equation F=ma. I that is the moment of inertia is like m in the linear case. The magnitude of a torque is defined as

$\tau = Fr\sin \theta$

with F the force applied in some point, r the distance of the point respect the axis rotation and θ the angle between the force and the radial vector that points toward the point the force is applied, in our case θ=90 and sinθ=1, then (1):

$Fr = I\alpha$ (2)

Because the applied force is constant the angular acceleration is constant too, and for constant angular acceleration we have that it's equal to the change of angular velocity over a period of time:

$\alpha=\frac{0.800}{2.00}=0.40 \frac{rev}{s^{2}}$

It's important to work in radian units so knowing that $1rev=2\pi rad$

$\alpha=2.51 \frac{rad}{s^{2}}$ (3)

The moment of inertia of a disk is:

$I=\frac{MR^{2}}{2}$ (4)

with M the mass of the disk and R its radius, then

$I=\frac{(140)(1.50)^{2}}{2}=157.5 kg*m^2$

using the values (3) and (4) on (2)

$Fr = (157.5)(2.51)$ (2)

Because the force is applied about the rim of the disk r=R=1.50:

$F= \frac{(157.5)(2.51)}{1.50}=263.55 N$