Circular Motion and Linear Speed

[Peter G.]

Hi,

Points P and Q are at distances R and 2R respectively from the centre X of a disc

The disc is rotating about an axis through X, normal to the plane of the disc. Point P has linear speed v and centripetal acceleration a. Which one of the following is correct for point Q?

Linear Speed Options:
v
v
2v
2v

Centripetal Acceleration Options:
a
2a
2a
4a

So, I tried to play around with: a = v² / r

From that I thought that as we double the Radius, as for Q, 2r we half the acceleration. But there's no option for that...

I am confused can anyone guide/hint me so I know what path to follow?

Thanks,
Peter G.

 

[PhanthomJay]

If you double the radius, the acceleration is halved only if the linear speed remains the same. The linear speed of point Q is not the same as the linear speed of point P.

I don't know why you have repeated your option choices.

 

[Fewmet]

From that I thought that as we double the Radius, as for Q, 2r we half the acceleration. But there's no option for that...

You are overlooking that the linear speed varies with distance from the center of rotation. It is given as V at R. Do you see how to find it at Q?

 

[Peter G.]

It is a multiple choice question, so, for option c, it would be: Linear Speed 2v and Centripetal Acceleration 2a.

And, from your explanations, I should therefore substitute v² for 4 x pi² x r / T²?

Thanks once again,
Peter G.

 

[Fewmet]

It is a multiple choice question, so, for option c, it would be: Linear Speed 2v and Centripetal Acceleration 2a.

And, from your explanations, I should therefore substitute v² for 4 x pi² x r / T²?.

Like PhatomJay, I'm a little unclear about the choices offered for each question.

Yes: the velocity out at 2R will be 2V.

I'm not sure why you are introducing $$pi$$. The acceleration out at 2R is found by substituting 2V and 2R into a = v$$^{}^{}2$$/r.

Do you see how that gives you twice the acceleration at P?

(Sorry about the sloppy formatting...I'm learning...)

 

[Peter G.]

So, regarding the choices offered:

It is a table, with two colums, four rows.
The heading on the columns are: Linear Speed and Centripetal Acceleration;
Option A: Linear Speed v, Centripetal Acceleration a
Option B: Linear Speed v, Centripetal Acceleration 2a
Option C: Linear Speed 2v, Centripetal Acceleration 2a
Option D: Linear Speed 2v, Centripetal Acceleration 4a

I'm just not being able to visualize how inputting 2R and 2V would output 2A and how would I know it is 2V before hand Sorry, can you show it maybe in more steps?

Thanks,
Peter G.

 

[Fewmet]

Hmm...It seems to me the answer is that the linear velocity at Q is 2V and the centripetal acceleration is 2a

Here is my reasoning:
Circumference at P = 2$$\pi$$R
Linear velocity at P =2$$\pi$$R/T = v (given)
Acceleration at P =v$$^{}2$$/R = a (given)
Circumference at Q = 2$$\pi$$(2R)
Linear velocity at Q = 2$$\pi$$(2R)/T=2v
Acceleration at Q = (2v)$$^{}2$$/2R = 2a

Could it just be that you are skipping steps and making an algebra mistake in the process?

I've skipped steps in the linear velocity and acceleration at Q because the formatting is behaving oddly at the moment).

 

[Peter G.]

Ah ok, now I understood. I was also thinking that the acceleration would be 4a because I was again thinking the radius was constant...

Thanks a lot for your patience