Uniform Circular Motion — Question 10

Question 10

An inward pull can require outward static friction

Take $g=10$ m/s$^2$. A 1 kg puck remains fixed relative to a horizontal rotating platform at radius 0.50 m. A central light cord is kept at constant tension 2 N by an ideal actuator. The static coefficient is 0.30. At each tested angular speed the motion is steady; ignore the spin-up process.

1. Find signed inward friction as a function of angular speed.

2. Find its magnitude and direction at $\omega=0$, 2 and 3 rad/s.

3. Find the full nonnegative range of angular speeds permitting the puck to remain fixed on the platform.

Question 10 - Solution

An inward pull can require outward static friction

(a) Use the radial force deficit.

Taking inward friction as positive, radial force balance gives $$T+f=m\omega^2r\Longrightarrow\boxed{f=.50\omega^2-2}\quad\text{(SI values)}.$$ Vertical balance supplies $N=mg=10$ N, hence $|f|\leq\mu_sN=3$ N.

(b) Allow the sign to reverse.

Substitute the three angular speeds: $$\boxed{f(0)=-2\ \mathrm N},\qquad\boxed{f(2)=0},\qquad\boxed{f(3)=+2.5\ \mathrm N}.$$ The first force is outward, the second vanishes, and the third is inward. Friction offsets the difference between cord tension and the required inward resultant.

(c) Impose the two-sided capacity.

Impose the full signed interval: $$-3\leq.50\omega^2-2\leq3 \Longrightarrow-2\leq\omega^2\leq10\ \mathrm{s^{-2}}.$$ Since $\omega\geq0$, the lower inequality is automatic: $$\boxed{0\leq\omega\leq\sqrt{10}=3.162\ \mathrm{rad/s}}.$$ These are separate steady states; spin-up forces were explicitly excluded.