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A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.05 s?

User JaredPar
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1 Answer

2 votes

Answer:

The angular speed is 0.13 rev/s

Step-by-step explanation:

From the formula


\tau = I\alpha

Where
\tau is the torque


I is the moment of inertia


\alpha is the angular acceleration

But, the angular acceleration is given by


\alpha = (\omega)/(t)

Where
\omega is the angular speed

and
t is time

Then, we can write that


\tau = (I\omega)/(t)

Hence,


\omega = (\tau t)/(I)

Now, to determine the angular speed, we first determine the Torque
\tau and the moment of inertia
I.

Here, The torque is given by,


\tau = rF

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N


\tau = 3.00 * 195


\tau = 585 Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by


I = (1)/(2)MR^(2)

Where M is the mass and

R is the radius


I = (1)/(2) * 325 * (3.00)^(2)


I = 1462.5 kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,


\omega = (\tau t)/(I)


\omega = (585 * 2.05)/(1462.5)


\omega = 0.82 rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

User Satej S
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6.6k points