Angular velocity, acceleration. and moment of inertia

• bfusco
In summary: well α=34.05 rad/sec^2, but i have no clue what the information the slowing of the wheel is giving, i want to know what the formula is for the applied torque needed to achieve a given angular velocity
bfusco

Homework Statement

A grinding wheel is a uniform cylinder with a radius of 8.70 cm and a mass of 0.400 kg.
a)calculate the moment of inertia about the center.
b)Calculate the applied torque needed to accelerate it from rest to 1950 rpm in 6.00 s if it is known to slow down from 1250 rpm to rest in 57.5 s.

The Attempt at a Solution

first i changed the given radius to .087 m..

a) wasnt that hard, I=mR^2, (.4)(.087)^2=.00151

b) since i was given the ωi=0, and ωf=1950 rpm→32.5 rps, and t=6s, i used the formula ωf=ωi+αt, and got α=5.42. i used my α to solve for a with the equation a=Rα (because F=ma) and got .47, then using F=ma, F=(.4)(.47)=.19. Since τ=RF, τ=(.19)(.087)=.02, which is incorrect.

Im guessing that my calculated I has something to do with the answer because of the way my professor asks the question, in parts where you use your answer in part a to solve for part b which is needed to solve for part c, etc, but i don't understand how my answer is wrong. however realistically i can understand how a τ of .02 wouldn't cause a cylinder to rotate from 0 to 1950 rpm in 6 sec.

Your acceleration should be expressed in unit of rad/s2 in order to be consistent with other calculations. You should ALWAYS attach units to your values.

You don't need to convert back and forth from torque to force via the radius. You can work directly with torques, keeping everything "angular"

Why do you think they gave you the information about the wheel slowing to a halt over a given time? What would bring something in motion to a halt over time?

gneill said:
Your acceleration should be expressed in unit of rad/s2 in order to be consistent with other calculations. You should ALWAYS attach units to your values.

You don't need to convert back and forth from torque to force via the radius. You can work directly with torques, keeping everything "angular"

Why do you think they gave you the information about the wheel slowing to a halt over a given time? What would bring something in motion to a halt over time?

i don't understand what you mean by "You can work directly with torques, keeping everything "angular"" in order to get torque i need the force, in order to get the force i need acceleration, so the only way to take what was given and connect it to acceleration is through the equation a=rα and α is calculated from the equation i used with ωf and ωi

bfusco said:
i don't understand what you mean by "You can work directly with torques, keeping everything "angular"" in order to get torque i need the force, in order to get the force i need acceleration, so the only way to take what was given and connect it to acceleration is through the equation a=rα and α is calculated from the equation i used with ωf and ωi

The entire problem is dealing with angular quantities; there are no linear references and none are required. You can do everything with torque, angular acceleration, and time. All of the "usual" linear kinematic equations have their angular counterparts. Thus ##\tau = I \alpha## is the angular equivalent to F = ma, for example.

gneill said:
The entire problem is dealing with angular quantities; there are no linear references and none are required. You can do everything with torque, angular acceleration, and time. All of the "usual" linear kinematic equations have their angular counterparts. Thus ##\tau = I \alpha## is the angular equivalent to F = ma, for example.

ok so then, τ=(.00151)(5.42)=.0082 kg*m^2*rev/sec? still sounds like to small of a number, and i am unsure of the units

bfusco said:
ok so then, τ=(.00151)(5.42)=.0082 kg*m^2*rev/sec? still sounds like to small of a number, and i am unsure of the units

I think that one problem is that you're mixing units inapropriately. Do your calculations in radians per second for angular velocities, and radians per second2 for angular accelerations.

5.42 is your value for revolutions per second2. Convert that to radians per second2.

Also, you haven't yet dealt with the information concerning the slowing of the wheel from some initial angular velocity to rest. What information does this give you?

gneill said:
I think that one problem is that you're mixing units inapropriately. Do your calculations in radians per second for angular velocities, and radians per second2 for angular accelerations.

5.42 is your value for revolutions per second2. Convert that to radians per second2.

Also, you haven't yet dealt with the information concerning the slowing of the wheel from some initial angular velocity to rest. What information does this give you?

well α=34.05 rad/sec^2, but i have no clue what the information the slowing of the wheel is giving, i want to say drag, but no i highly doubt it

bfusco said:
well α=34.05 rad/sec^2, but i have no clue what the information the slowing of the wheel is giving, i want to say drag, but no i highly doubt it

Another word for drag is friction...

So there's a frictional torque acting.

1. What is angular velocity?

Angular velocity is a measure of the rate of change of angular displacement, typically measured in radians per second. It describes how quickly an object is rotating around an axis.

2. How is angular velocity related to linear velocity?

Angular velocity and linear velocity are related through the equation v = rω, where v is linear velocity, r is the distance from the axis of rotation, and ω is angular velocity. This means that as angular velocity increases, so does linear velocity.

3. What is angular acceleration?

Angular acceleration is the rate of change of angular velocity over time. It is typically measured in radians per second squared and describes how quickly the angular velocity is changing.

4. How is angular acceleration related to torque?

Angular acceleration is directly proportional to torque, which is a measure of the force applied to rotate an object around an axis. This relationship is described by the formula α = τ/I, where α is angular acceleration, τ is torque, and I is the moment of inertia.

5. What is moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is calculated by taking the mass of the object and the square of its distance from the axis of rotation into account. Objects with a larger moment of inertia require more torque to achieve the same angular acceleration as objects with a smaller moment of inertia.

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