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Logindynamics
12 posts
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Just to get you started.
Q A)
v=rω
and ω is the same for both small and large discs (they have to roll at the same angular velocity)
so ω = v/r = 2.2/0.388
so v at pt A = 2.2/0.388*0.717 = 4.065 m/s
a at pt A = v^2 / r = 4.065^2/0.717 = 23.05 m/s^2
The rest I do when I finish tuition. Rushing out to give now.
Edited by eagle 06 Sep `08, 9:36AM
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Originally posted by eagle:
Just to get you started.
Q A)
v=rω
and ω is the same for both small and large discs (they have to roll at the same angular velocity)
so ω = v/r = 2.2/0.388
so v at pt A = 2.2/0.388*0.717 = 4.065 m/s
a at pt A = v^2 / r = 4.065^2/0.717 = 23.05 m/s^2
The rest I do when I finish tuition. Rushing out to give now.
not the right answer =/ do u have to take measurements from the "C" point? the point which is in contact with the ground at that instant.
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Originally posted by eagle:
paiseh... see many things wrong.... quite hard to rotate between the graphics and typing becuz I type the answer out direct -.-"
try again
angular acceleration = 5.7/0.194 (radius) = 29.38
So, linear acceleration at point A = 29.38 * 0.3585 = 10.5 m/s^2
still no luck :l this question is soo hard we tried soo many different things and still get it wrong. hmmm... somehow we should use all the info they give us
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hmmm... long time since I did solid body...
Still think the 2nd try for Q A is correct leh... Not necessary must use all value I think. Because v^2/r doesn't give you a => The discs are still accelerating... not just rotating at a constant speed.
Anyway, Q B
Linear acc component perpendicular to B is 235 mm/s^2 sin 32 = 124.53 mm/s^2
We need to find the centre of rotation. So 124.53 / 0.2312 = 538.6 mm from B (in the direction of A for the angular direction as shown)
So A is 538.6 - 371 = 167.6 mm from the centre of rotation.
Magnitude of acceleration at A = 0.2312 * 167.6 = 38.75 mm/s^2
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