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just olya [345]
4 years ago
11

Rotational dynamics about a fixed axis: A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through

its center. Starting with an angular velocity of 2.40 rev/s, it stops after turning through 18.2 rev with uniform acceleration. The net torque acting on this sphere as it is slowing down is closest to:A) 0.149 N m. B) 0.0620N m. C) 0.00593 N m. D) 0.0372 N m. E) 0.0466 N·m
Physics
2 answers:
solong [7]4 years ago
6 0

Answer:

(D) The net torque acting on this sphere as it is slowing down is closest to  0.0372 N.m

Explanation:

Given;

mass of the solid sphere, m =  1.85 kg

radius of the sphere, r = ¹/₂ of diameter = 22.5 cm

initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s

angular revolution, θ = 18.2 rev = 114.37 rad

Torque on the sphere, τ = Iα

Where;

I is moment of inertia

α is angular acceleration

Angular acceleration is calculated as;

\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)

moment of inertia of solid sphere, I = ²/₅mr²

                                                           = ²/₅(1.85)(0.225)²

                                                           = 0.03746 kg.m²

Finally, the net torque on the sphere is calculated as;

τ = Iα

τ = 0.03746 x 0.994

τ = 0.0372 N.m

Therefore, the net torque acting on this sphere as it is slowing down is closest to  0.0372 N.m

zubka84 [21]4 years ago
5 0

Answer:

D) 0.0372 N m

Explanation:

r = 45/2 cm = 22.5 cm = 0.225 m

As 1 revolution = 2π rad we can convert to radian unit

2.4 rev/s = 2.4 * 2π = 15.1 rad/s

18.2 rev = 18.2 * 2π = 114.35 rad

We can calculate the angular (de)acceleration using the following equation of motion

-\omega^2 = 2\alpha \theta

- 15.1^2 = 2*\alpha * 114.35

\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2

The moment of inertia of the solid uniform sphere is

2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2

The net torque acting on this according to Newton's 2nd law is

T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm

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