dynamics

• WoAiMeiMei

  6 Sep 08, 09:19

  help :O i need to get this correct by monday and so far i havent had any luck with my friend either >_< i dont understand

  

   

  :l

  http://i105.photobucket.com/albums/m225/nickykeeng/album2/jf3.jpg

• eagle

  6 Sep 08, 09:36

  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.

• WoAiMeiMei

  6 Sep 08, 09:50

  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.
• eagle

  6 Sep 08, 11:59

  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

   

• WoAiMeiMei

  6 Sep 08, 12:25

  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
• eagle

  6 Sep 08, 20:34

  what are the given answers?

• WoAiMeiMei

  6 Sep 08, 22:31

  i dun have.. i wish i did

• eagle

  7 Sep 08, 09:16

  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

   

• eagle

  7 Sep 08, 09:26

  Q C

  The answer should still be 4.3 m/s^2 I think

  Anyway, how do you check whether it is correct or wrong?

• WoAiMeiMei

  7 Sep 08, 11:53

  we chuck it online on a program. ill look at those 2 tomorrow when meet my friend either tonight or tomorrow. :O
• tr@nsp0rt_F3V3R

  22 Sep 08, 23:38

  sry... but i think... and i thought... this chapter is no longer in the o's?

• eagle

  23 Sep 08, 00:02

  erm... TS is a uni student