Solve Rotational Motion: Angular Acceleration, Torque & Force

AI Thread Summary
A wheel decelerates from 100 rpm to a stop in 10 seconds, leading to a calculation of angular acceleration, torque, and force. The angular acceleration was initially miscalculated but clarified to be 1.05 rad/sec². Torque was determined to be 0.00016 kgm², and the force exerted by the torque on a fixed object was calculated using the formula Torque = R x F, resulting in a force of 0.064 N. The discussion emphasized the distinction between angular and linear acceleration and confirmed the correct application of torque formulas. Overall, the calculations and understanding of the concepts were validated through collaborative input.
tupldy
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Homework Statement



A wheel decellerates from 100 rpm in 10 sec. Find angular acceleration, torque and the force torque is producing on a fixed object.
The wheel has a radius of .0025m and a mass of 50kg.

Homework Equations


T=Ia
a=ra
T=dL/dt


The Attempt at a Solution



The first thing I did was change rpm to rad/sec which gave me 10.5 rad/sec.
for angular acceleration i used: a=ra-> .0025*10.5/10 which gave me
a=.002625 rad/sec^2.
For torque i used 1/2*50*.0025^2*1.05 which gave me .00016 kgm^2.
I am not sure where to go with this now, or even if I am headed in the right direction. Any suggestions would be greatly appreciated.
 
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You have not correctly found the angular acceleration. The definition of angular acceleration is;

\alpha = \frac{d\omega}{dt}
 
wouldn't the dw in this case just be the 10.5 rad/sec divided by 10(then number of seconds it takes for the wheel to stop?
 
tupldy said:
a=ra
I assume you mean:
a = r \alpha
This relates angular acceleration to linear acceleration.
The first thing I did was change rpm to rad/sec which gave me 10.5 rad/sec.
OK.
for angular acceleration i used: a=ra-> .0025*10.5/10 which gave me
a=.002625 rad/sec^2.
You are mixing up angular and linear acceleration. All you need is angular acceleration.
 
yes, your formula for acceleration is the one i meant. I understand that dw/dt is angular acceleration, so do i just not multiply by the radius? I am very confused here.
 
tupldy said:
I understand that dw/dt is angular acceleration, so do i just not multiply by the radius?
Correct. You would only multiply by the radius if you needed the linear acceleration (units: m/s^2) of a point on the rim of the wheel.
 
ok so now that i know torque of the wheel is .00016kgm^2 and angular acceleration is 1.05 rad/sec^2. How do i figure out what the force is that the torque of the wheel is producing on a fixed object, say someone's head for example?
 
If you assume that the torque is created by a force applied tangentially to the edge of the wheel, what force is required to generate that torque? (What's the definition of torque?)
 
Torque = R xF or Torque = I alpha or Torque = dL/dt.
I understand that these are the formulas for torque, I am just not sure which one I am supposed to use.
 
  • #10
I decided to try T=R x F. What I got was that the torque of the wheel is equal to the radius times the unknown F times the sin90.

.00016=.0025F

F=.064N
Does this look right to you?
 
  • #11
Well, what are you trying to find? Which formula relates to that quantity?
 
  • #12
I am trying to find the force the torque exerts on the object The formula I used was Torque = R x F.
 
  • #13
tupldy said:
I decided to try T=R x F.
Wise choice, since that's the only one that has force in it. :wink:

What I got was that the torque of the wheel is equal to the radius times the unknown F times the sin90.

.00016=.0025F

F=.064N
Does this look right to you?
Looks OK to me.
 
  • #14
Great thank you so much for all your help. :0)
 
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