Two 1.30 kg balls are attached to the ends of a thin rod of length 30.0 cm and n

Question

Two 1.30 kg balls are attached to the ends of a thin rod of length 30.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (the figure), a 36.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 2.34 m/s and then sticking to it. (a) What is the angular speed of the system just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle (deg) will the system rotate before it momentarily stops?

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A top spins at 33 rev/s about an axis that makes an angle of 28° with the vertical. The mass of the top is 0.54 kg, its rotational inertia about its central axis is 4.3 x 10-4 kg·m2, and its center of mass is 3.7 cm from the pivot point. The spin is clockwise from an overhead view. (a) What is the precession rate? (b)What is the direction of the precession as viewed from overhead?

Explanation / Answer

Q1.

a)
The angular momentum Li of the system before the collision is due to the wad.
Li = rMv, where r = l/2 = 15 cm = 0.15m

Lf = If*w = (mr^2+mr^2+Mr^2)w
= (2m+M)r^2*w

So, Angular velocity w = rMv/(2m+M)r^2
=> w = Mv/(2m+M)r
=> w = (0.036*2.34)/((2*1.30+0.036)*0.15)
=> w = 0.213 rad/s

b)
Initial KE Ki = Mv^2/2

Final KE Kf = If*w^2/2
= M^2*v^2 / 2(2m+M)

So,
Kf/Ki = [(M^2*v^2) / (2(2m+M))]*[2/Mv^2]
= M/2m+M
= 1/(1+2m/M)
= 1/(1+(2.6/0.036))
= 0.0136

c)
sin = h/Rcom = [M^2*v^2 / 2(2m+M)^2*g]*[(2m+M)/Mr]
= Mv^2/2(2m+M)gr
= (0.036*2.34^2)/(2(2.6+0.036)*9.8*0.15)
= 0.02543557
So,
= 180 + sin^-1(0.02543557)
= 181.4 deg

Please post other question separately.

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