
#1
Sep2005, 03:43 AM

P: 39

A puck of mass 80.0 g and radius 4.20 cm slides along an air table at a speed of 1.50 m/s. It makes a glancing collision with a second puck of radius 6.00 cm and mass 140.0 g (initially at rest) such that their rims just touch. Because their rims are coated with instantacting glue, the pucks stick together and spin after the collision.
How do i find the angular momentum of the system relative to the centre of the mass? I have found the centre of mass already. It's 10.7cm from the perimeter of the small puck. 



#2
Sep2005, 04:19 AM

P: 68

Firstly
Consider the distance of the centre of gravity from the centre of the first (moving puck) that might help 



#3
Sep2205, 09:33 AM

P: 39

I have already calculated that distance and it is 10.7 cm from the small puck! From here, I used the formula angular velocity=linear velocity/ radius = 1.5 / 0.107 =14.0 rad/s. From here I calculated the moment of inertia of the 2 pucks = 0.0008522. Thus I will get the angular momentum which is the product of the two answers above = 0.0119. Is this the right answer?




#4
Sep2205, 11:49 AM

Mentor
P: 40,905

HELP! Urgent on question of centre of mass and parallel axis theorem 



#5
Sep2705, 03:53 PM

P: 39

Ermm... ... any help in the first steps for the first particle? I try doing the second one by myself. Just help me out with the first particle.. Thanx!




#7
Sep2905, 04:31 AM

P: 39

talking abt the linear speed right???




#8
Sep2905, 07:45 AM

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P: 40,905

That's right. If you want to find the angular momentum with respect to the center of mass, you'll need to measure the speed of each puck with respect to the center of mass. But first you should find the speed of the center of mass itself.




#9
Oct105, 09:10 AM

P: 39

but u mean finding the linear speed abt the centre of mass before the 2 bodies even get stuck or what? cos their centre of mass is different as only 1 particle was moving initially...




#10
Oct105, 02:03 PM

P: 39

Pls Help@ Urgent! Thanx!




#11
Oct105, 05:41 PM

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P: 40,905

Yes, find the speed of the center of mass before the collision. (Which is the same as the speed of the center of mass after the collision, since no external force acts on the system.)




#12
Oct205, 03:26 AM

P: 39

Since only 1 particule is moving, the linear KE = 1/2 mv^2. But again, who do we find the sped of the centre of mass. At this pt in time, the centre of mass of the SYSTEM isn't moving. it only moves rotationally after itwas hit by the small puck. So wldnt the speed be zero? only the small puck has the speed it needed.




#13
Oct205, 07:37 AM

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P: 40,905

The center of mass of the system is most definitely moving! If both pucks were stationary, then the center of mass would not move. But since the center of mass of the system depends on the position of both pucks, as one puck moves the center of mass also moves.
Big hint: Horizontal position of center of mass: [tex]M_{total} x_{cm} = m_1 x_1 + m_2 x_2[/tex] Horizontal speed of center of mass: [tex]M_{total} v_{cm} = m_1 v_1 + m_2 v_2[/tex] 



#14
Oct205, 01:24 PM

P: 39

ur x here is measured relative to?




#15
Oct205, 01:52 PM

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P: 40,905

x is the horizontal position relative to whatever origin you like, as long as it's fixed with respect to the table.




#16
Oct605, 01:52 PM

P: 39

Oh yeah I got that thks! But its really darn long.... .. :(



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