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Alenkasestr [34]
2 years ago
3

A thin, cylindrical rod ℓ = 27.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attach

ed to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.
(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy?


(b) What is the angular speed of the rod and ball?


(c) What is the linear speed of the center of mass of the ball?


(d) How does it compare with the speed had the ball fallen freely through the same distance of 32.0 cm?
Physics
1 answer:
Anuta_ua [19.1K]2 years ago
7 0
(a) PE is converted into KE. Measured relative to the pivot point, initial
PE = (1.20kg * 0.270m + 2.00kg * (0.0500 + 0.270)m) * 9.8m/s²
PE = 9.45 J ◄ becomes the KE at 90º

(b) KE = 9.45 J = ½Iω²
where
I = I_rod + I_sphere
I = mL²/3 + (2/5)MR² + M(R+L)² → assuming the ball is solid
and invoking the parallel axis theorem
I = 1.20kg * (0.270m)² / 3 + (2/5) * 2.00kg * (0.0500m)² + 2.00kg * (0.320m)²
I = 0.236 kg·m²

so
9.45 J = ½ * 0.236kg·m² * ω²
ω = 8.95 rad/s ◄ angular speed of rod and ball

(c) v = ωr = 8.95rad/s * (0.270 + 0.0500)m = 2.86 m/s ◄

(d) V = √(2gh) = √(2 * 9.8m/s² * 0.320m) = 2.50 m/s ◄
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Answer:

1.4 *10^6 N/C

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