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julsineya [31]
3 years ago
6

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 a

bout 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
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

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

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