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

In summary, the conversation discusses the given scenario of a wheel rotating on a vertical plane and its angular acceleration, starting from rest. The question asks for the angular acceleration, tangential speed, total acceleration, and angular position of a point on the wheel at t=2sec. The equations needed to solve these are also mentioned. The conversation also touches on approximating degrees in radians and the direction of rotation of the wheel. The conversation ends with a suggestion to use calculus for solving the angular position or to use the kinematic equation.
  • #1
genioaddis
3
0

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:
 
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  • #2
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.
 
  • #3
Thanks @mafagafo. I see what you mean. But one last question, which way is the wheel turning?
 
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  • #4
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.
 
  • #5
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 waht 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)
 
  • #6
Oh wow... Forgot to post the diagram XD
 

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  • #7
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.
 

1. What is a wheel on a vertical plane?

A wheel on a vertical plane refers to a wheel that is positioned vertically, meaning the axle is perpendicular to the ground. This is different from a horizontal plane where the wheel's axle is parallel to the ground.

2. How do you calculate tangential speed of a wheel on a vertical plane?

The tangential speed of a wheel on a vertical plane can be calculated by multiplying the radius of the wheel by its angular velocity. This is because the tangential speed is the linear distance traveled by a point on the wheel's circumference in a given amount of time.

3. What factors affect the tangential speed of a wheel on a vertical plane?

The tangential speed of a wheel on a vertical plane is affected by the radius of the wheel and its angular velocity. The larger the radius or the faster the wheel's rotation, the greater the tangential speed will be. Other factors that may affect the tangential speed include friction and air resistance.

4. How is acceleration calculated for a wheel on a vertical plane?

The acceleration of a wheel on a vertical plane can be calculated by dividing the change in tangential speed by the change in time. This is because acceleration is the rate of change of an object's velocity over time. Therefore, a change in tangential speed will result in a change in acceleration.

5. Why is it important to consider the tangential speed and acceleration of a wheel on a vertical plane?

The tangential speed and acceleration of a wheel on a vertical plane are important to consider because they can affect the stability and performance of the wheel. For example, a higher tangential speed can make the wheel more likely to slip or lose traction, while a higher acceleration can cause the wheel to spin out of control. Understanding these factors can help in designing and optimizing the performance of wheels on vertical planes, such as those used in amusement park rides or bicycles.

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