Wheel on a vertical plane. Find tangential speen and acceleration.

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SUMMARY

The discussion focuses on calculating the tangential speed, angular position, and total acceleration of a wheel rotating in a vertical plane, given an angular acceleration of 4 rad/sec² and a radius of 1m. At t=2 seconds, the angular velocity is determined to be 8 rad/sec, while the tangential speed can be calculated using the formula v=ωr. The participants emphasize the importance of using radians instead of degrees for accurate calculations and clarify that gravity does not exert torque on the wheel's center of mass.

PREREQUISITES
  • Understanding of angular motion and kinematics
  • Familiarity with the equations of motion: ωf=αt + ωo, v=ωr
  • Knowledge of trigonometry for calculating tangential speed
  • Basic calculus concepts for angular position integration
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  • Learn how to apply angular kinematics in vertical motion scenarios
  • Study the effects of gravity on rotating bodies
  • Explore the relationship between angular velocity and tangential speed in detail
  • Investigate torque and its impact on rotational dynamics
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Students studying physics, particularly those focusing on rotational motion, as well as educators and anyone interested in understanding the dynamics of objects in vertical planes.

genioaddis
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Homework Statement



Hey all! I'm new here but I think I'll like it. :) I have a question that I can't seem to figure out. Here is the question in short:
GIVEN: A wheel of radius 1m on a vertical plane rotates with an angular acceleration of 4rad/sec2. The wheel starts from rest (ωo = 0). Point P, on the wheel, is at 57.3° from the horizontal.
REQUIRED: At t=2sec, (a) what is the angular acceleration of the wheel? (b) What is the tangential speed and the total acceleration at P? and (c) What is the angular position of P?

Homework Equations



Some equations needed might be:
ωf=αt + ωo
v=ωr
θ=s/r

The Attempt at a Solution


I have attempted all of them. For (a) my answer is to use the first formula I wrote to find 8rad/sec. For (b) I am confused because I don't know if the rotation is Clockwise or Counter Clockwise. Since the rotation is vertical, gravity must have some role in it. Does gravity help the rotation at P or is it against it? (c) I think this one is easy. Solve for s in the third formula. 57.3 radians is my answer.

Please help me out. Thanks so much! :smile:
 
Last edited:
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genioaddis said:
angular acceleration of 4rad/sec.

To start off, this is not an acceleration. (The units are wrong.)

genioaddis said:
at 57.3° from the horizontal.

Can you approximate this value in radians?

genioaddis said:
ωf=αt + o
What is this?

genioaddis said:
v=ωr
Right.

genioaddis said:
θ=s/r
Right. Here, if you use degrees, you will get that the circumference is equal to the product of the radius by 360°. SEE? Use radians. Reread this last sentence.

genioaddis said:
Does gravity help the rotation at P or is it against it?
Think about it. If it is clockwise, what does that imply? If it is not, what can you conclude from that? What is the gravity force doing in one side of the wheel? And in the other?

genioaddis said:
57.3 radians is my answer.
Simply redo your work and PICK EITHER degrees or radians.
 
Thanks @mafagafo. I see what you mean. But one last question, which way is the wheel turning?
 
Last edited:
I don't like to give answers. You are supposed to learn from here (and this makes this forum so awesome).

If you still have questions, ask them. Don't be ashamed.

If it had an angular velocity of 4 rad/s, the wheel was turning... (go on your own, TIP: wikipedia has it).

It is a standard, as you may have supposed. The negative velocity (-4 rad/s) indicates that the wheel is moving in the other direction.
 
First off, Wecome to PF. I think you will like it here =]

gravity acts through the center of mass. Therefore it doesn't exert a torque on an object that is rotating about its center of mass (r=0 thus rxF=0)
Is it's acceleration 4rad/sec^2? If that's the case, given the information above, I don't see a reason why that would change, do you? Are there any torques acting on the wheel? your first equation is "good", but I think you meant ##\omega_{f} = \alpha t + \omega_{0}##
I'm not sure what you're 3rd equation is, is that some kind of trig representation? Or what is 's'?

Anyway, for (b), for the tangential speed, you can use trig to figure that out. See the diagram, I think you can figure it out.
(c) Do you know calculus? ##\theta (t) = \int \omega (t)dt## If not, use the kinematic they gave you in class (the calculus is done for you)
 
Oh wow... Forgot to post the diagram XD
 

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Thanks! I see gravity has no effect. I thought of it as a ball rotating on string.
BiGyElLoWhAt said:
Or what is 's'?
's' the linear distance.
 

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