Question

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

Answer 1

Answer:

The net torque is 0.0372 N m.

Explanation:

A rotational body with constant angular acceleration satisfies the kinematic equation:

$\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta$ (1)

with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

$\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha$

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

$\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}}$

The negative sign indicates the sphere is slowing down as we expected.

Now with the angular acceleration we can use Newton's second law:

$\sum\overrightarrow{\tau}=I\overrightarrow{\alpha}$ (2)

with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:

$I = \frac{2MR^{2}}{5}$

With M the mass of the sphere an R its radius, then:

$I = \frac{2(1.85)(\frac{0.45}{2})^{2}}{5}=0.037 kg*m^2$

Then (2) is:

$\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm$

Answer 2

Answer:

D. 0.037 N m.

Explanation:

Given:

Diameter, d = 45 cm

Radius, r = d/2

= 0.225 m

Mass, m = 1.85 kg

Initial velocity, wi = 2.4 rev/s

θ = 18.2 rev

Final velocity, wf = 0 rev/s

Inertia of a uniform sphere, I = 2/5 × m × R^2

= 2/5 × 1.85 × 0.225^2

= 0.0375 kg.m^2

Using equation of angular motion,

wf^2 = wi^2 + 2ao × θ

Where,

ao = angular acceleration

ao = (2.4^2)/2 × 18.2

= 0.158 rev/s^2

Converting from rev/s^2 to rad/s^2,

0.158 rev/s^2 × 2pi rad/1 rev

= 0.993 rad/s^2

Torque, t = I × ao

= 0.0375 × 0.993

= 0.037 Nm