A puck of mass 80.0 g and radius 4.10 cm slides along an air table at a speed of

Question

A puck of mass 80.0 g and radius 4.10 cm slides along an air table at a speed of 1.50 m/s as shown in Figure P11.34a. It makes a glancing collision with a second puck of radius 6.00 cm and mass 125.0 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision (Fig. P11.34b). (a) What is the angular momentum of the system relative to the center of mass? (123/25000) Your response differs from the correct answer by more than 10%. Double check your calculations. kg middot m^2/s (b) What is the angular speed about the center of mass? rad/s

Explanation / Answer

Given:

m1= 80g = 0.08kg

m2=125g = 0.125kg

M = m1 + m2 = 205g = 0.205kg

co- ordinates of mass m1 = (x1 , 10.1) [10.1cm= 4.1cm+6cm]

co- ordinates of mass m2 = (0,0)

Solution:
Let the center of mass of the puck of the 125 g(m2 let) be at the origin before the collision.

(a) Before the collision,

the y-coordinate of the CM is
(m1y1 + m2y2)/M = (0.08*0.11 +0.205*0)/0.205 = 0.0088/0.205 = 0.0429
The x-coordinate of the CM is

(m1x1 + m2x2)/M = (0.80*x1+0.205*0)/0.205 = 0.80x1 /0.205 = 0.0429x1.
The velocity of the CM is
vCM = dxCM/dt

= (0.41) dx1/dt = (0.41)1.5 m/s = 0.615 m/s in the x-direction.
According to question,

the particle moves with velocity v = 1.5 m/si  
and the CM moves with vCM = 0.615 m/s i.
With respect to the CM m1 moves with velocity v1 = v - vCM = 1.5i - 0.615i = 0.885 i m/s
and m2 moves with velocity v2 = 0 - vCM = -0.615i m/s.
The angular momentum of the system about the CM is
L = -(m1v1(y1 - yCM) + m2v2yCM )k = -(0.80*0.885i(0.1-0) +0.205*(-0.615)*0)k = -7.08k = -(7.08*10-3kgm2/s2)k Ans

(b) In the collision momentum and angular momentum are conserved.

The total angular momentum about the CM after the collision is I = (7.08*10-3 kg m2/s )k.

The moment of inertia of the system about the CM is

I = I1 +  I2,

where I1 = ICM1 + M1R12, and I2 = ICM2+ M2R22.
I1 = (1/2) 0.08 kg(0.041m)2 + 0.08 kg(0.06 m)2 = 3.5524*10-4 kg m2.
I2 = (1/2) 0.125k g(0.06 m)2 + 0.12 kg(0.041 m)2 = 4.351*10-4 kg m2.
I = I1 + I2= 7.903*10-4kg m2.
= L/I = (8.962/s) k.
To find the moment of inertia of the composite object after the collision ,the theorem used is parallel axis theorem..

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