A puck of mass m1 = 60.0 g and radius r1 = 3.90 cm glides across an air table at

Question

A puck of mass m1 = 60.0 g and radius r1 = 3.90 cm glides across an air table at a speed of = 1.50 m/s as shown in Figure a. It makes a glancing collision with a second puck of radius r2 = 6.00 cm and mass m2 = loo g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and rotate after the collision (Figure b). (a) What ?s the magnitude of the angular momentum of the system relative to the center of mass? .0059 Your response is within 10 % of the correct value. This may be due to round off error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize round off error. kg .m^2/s (b) What is the angular speed about the center of mass? rad/s

Explanation / Answer

The distance between puck centers when they touch is 9.9 cm. The center of mass is located at a point along the line segment joining their centers and at such a distance that their moments of mass are equal: r1m1 = r2m2.

r1 + r2 = 9.9 cm, so that r2 = (9.9 cm - r1).

r1*60 g = r2*100 gm = (9.9cm - r1)*100 gm

160g*r1 = 990 gm*cm, and

r1 = (990 gm*cm)/(160 gm) = 6.1875 cm for the mass m1. This leaves 3.7125 cm for the mass m2.

In the initial case, where the pucks have not yet come together, the only angular momentum is that of the moving smaller puck. With a speed of 1.50 m/s, or 150 cm/s, it has an angular momentum of r1 times this speed and times its mass, m1, when taken around the center of mass:

Am1 = (150 cm/s)*(6.1875cm)(60gm)

Am1 = 55,687.5 (cm^2*gm/s) = 5.56875 [10^(-2)] (kg*m^2/s), since
1(cm^2*gm/s) = 10(-7)kg*m^2/s. This is the answer to part (a).

ans. is 5.56875 x 10-2 kg*m^2/s

The angular momentum will be the same after the collision as it was before it, but each mass will have only a part of this total angular momentum. The two masses will have the same angular speed around the center of mass after the collision: call this Va.
The angular speed equals the linear speed divided by the radius from the center or rotation, which is the center of mass in this case.
So after the collision:

m1*Va*r1^2 + m2*Va*r2^2 = 55,687.5(cm^2*gm/s)

[60 gm*(6.1875 cm)^2] + [100 gm*(3.7125 cm)^2]*Va = 55,687.5 (cm^2*gm/s)

[(2297.10 + 1378.26 )(g*cm^2)]*Va = (3675.36 gm*cm^2) = 55,687.5 (cm^2*gm/s) and thus

Va = [55,687.5(cm^2*gm/s)]/(3675.36 gm*cm^2) = 15.15 (radians)/s Answer to (b) part.

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