Physics Dynamics: Angular and Linear Motion of Machine Components

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Gear 'B' accelerates from rest to 250 rpm in 2 seconds, leading to an angular velocity of 26.2 rad/s and an angular acceleration of 13.1 rad/s². The tangential velocity at Gear 'B' is calculated to be 524 cm/s, with a tangential acceleration of 262 cm/s². For Gear 'A', the angular velocity is determined to be 34.9 rad/s and the angular acceleration 17.5 rad/s². The linear velocity and acceleration of rack 'C' are found to be 1047 cm/s and 525 cm/s, respectively, assuming uniform tooth spacing across the gears. The calculations are confirmed to be correct based on the provided parameters.
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Homework Statement



Gear ‘B’ on the machine shown below accelerates from rest to 250 rpm in 2 seconds. Determine the angular velocity and acceleration of A, and the linear acceleration and velocity of rack C. See image attached



Homework Equations



I don't know where i would start off with this question, help would be appreciated.


The Attempt at a Solution

 

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Welcome to PF!

tom23 said:
Gear ‘B’ on the machine shown below accelerates from rest to 250 rpm in 2 seconds. Determine the angular velocity and acceleration of A, and the linear acceleration and velocity of rack C.

Hi tom23! Welcome to PF!:smile:

(how far apart are the teeth on B compared with A and C? :confused:)

First find the angular velocity and acceleration of B.

Then you know that the "tooth-rates" on A B and C are the same … eg if B rotates through 7 teeth of the cog, then both A and C also move through 7 teeth. :wink:
 
So I got this until now, is it right?

ang velocity = 250 rpm x 2 pi x (1/60) = 26.2 rad/s

ang acceleration = change in ang velocity / chang in time = 26.2 rad/s / 2s = 13.1 rad/s^2

Thus at B: tangential Velocity at B = ang velocity*radius1 = 26.2*20 = 524 cm/s

tangential Acceleration at B = ang acceler.*radius1 = 13.1*20 = 262 cm/s^2

Therefore on A:

Angul. Velocity at A = tang. velocity/radius2 = 524/15 = 34.9 rad/s
Angul. Acceler at A = tang. acceler./radius2 = 262/15 = 17.5 rad/s

Does this two answers sound right, now how would i go on to solve for linear accel. and velocity of rack C?
 
tom23 said:
So I got this until now, is it right?

ang velocity = 250 rpm x 2 pi x (1/60) = 26.2 rad/s

ang acceleration = change in ang velocity / chang in time = 26.2 rad/s / 2s = 13.1 rad/s^2

Thus at B: tangential Velocity at B = ang velocity*radius1 = 26.2*20 = 524 cm/s

tangential Acceleration at B = ang acceler.*radius1 = 13.1*20 = 262 cm/s^2

Yes, that's fine. :smile:
Therefore on A:

Angul. Velocity at A = tang. velocity/radius2 = 524/15 = 34.9 rad/s
Angul. Acceler at A = tang. acceler./radius2 = 262/15 = 17.5 rad/s

You haven't answered my question: how far apart are the teeth on B compared with A and C?

If they're the same distance apart, then your answer is right.
now how would i go on to solve for linear accel. and velocity of rack C?

Same method … you have the tangential velocity of A, so you can find the linear velocity of C.
 
Tey didnt give the distance, so i believe it is the same, so for linear velocit and accel at C, i got:

V = 34.9 (30cm) = 1047 cm/s
a = 17.5 (30cm) = 525 cm/s

Does it look right?

Thanlks for all your help above!
 
Hi tom23! :smile:
tom23 said:
Tey didnt give the distance, so i believe it is the same, so for linear velocit and accel at C, i got:

V = 34.9 (30cm) = 1047 cm/s
a = 17.5 (30cm) = 525 cm/s

Does it look right?

Thanlks for all your help above!

Yes … that's fine … if the teeth are all the same distance apart, then the tangential velocities and the linear velocity are all the same. :smile:

(though I'd have done it directly from B … 26.2(40), rather than from A, as you did! :wink:)
 
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