Calculating Moment of Inertia and Torque for a Rotating Cylinder

AI Thread Summary
The discussion focuses on calculating the moment of inertia and torque for a grinding wheel, which is a uniform cylinder with a radius of 8.50 cm and a mass of 0.580 kg. The moment of inertia is calculated as I = 0.002 kg/m². To determine the applied torque needed to accelerate the wheel from rest to 1500 rpm in 5.0 seconds, the angular acceleration must first be calculated, factoring in the known deceleration from 1500 rpm to rest in 55.0 seconds. The final torque calculations yield a required torque of approximately 0.069 M/N when considering both the applied and resistive forces. Understanding the relationship between torque, moment of inertia, and angular acceleration is crucial for solving this problem effectively.
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Homework Statement


A grinding wheel is a uniform cylinder of with a radius of 8.50 cm and a mass of 0.580 kg. Calculate a) its moment of inertia about its center, and b) the applied torque needed to accelerate it from rest to 1500 rpm in 5.0 s if it is known to slow down from 1500 rpm to rest in 55.0 s.


Homework Equations


I=1/2(mr2)=.002kg/m2
\tau=I(\Deltaw/\Deltat)=300MN


The Attempt at a Solution

It doesn't seem that'' it is known to slow down from 1500 rpm to rest in 55.0 s'' ? Do i really need to use that information?
 
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if it is known to slow down from 1500 rpm to rest in 55.0 s.
May be there is a resistive force which causes it to slow down to rest by itself from 1500rpm in 55s. You might want to think about this deceleration.
 
Another for torque is torque = Inertia*angularAcceleration. I think you need to find angular acceleration first. acceleration = radius*alpha. Your rpm is rev/min convert this to rad/s to get omega(angular speed).

Some formulas:
v=rw
1 rev= 2pi

There are a couple of substitutions that you might need to do.
 
ok so i calculated the torque required disregarding the resistive force and got \tau=.063M/N then i calculated the torque of the resistive force and got \tau=-.006M/N so i figured i just add .006M/N of torque to .063 and got .069M/N
 
Assuming that your values are correct, you are right acc. to me.
 
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