Question

an ant of mass m clings to the rim of a flywheel of radius r, as shown above. the flywheel rotates clockwise on a horizontal shaft s with constant angular velocity ω. as the wheel rotates, the ant revolves past the stationary points i, ii, iii, and iv. the ant can adhere to the wheel with a force much greater than its own weight.

1. it will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points?

(a) i (b) ii (c) iii (d) iv
(e) it will be equally difficult for the ant to adhere to the wheel at all points.

1. what is the magnitude of the minimum adhesion force necessary for the ant to stay on the flywheel at point iii?

(a) mg (b) mω²r (c) mω²r² + mg (d) mω²r - mg (e) mω²r + mg

1. a figure of a dancer on a music box moves counterclockwise at constant speed around the path shown above. the path is such that the lengths of its segments, pq, qr, rs, and sp, are equal. arcs qr and sp are semicircles. which of the following best represents the magnitude of the dancers acceleration as a function of time t during one trip around the path, beginning at point p?

(a) graph (b) graph (c) graph (d) graph (e) graph

Explanation:

Question 10

Step1: Analyze centripetal force at each point

At point I, the centripetal force \(F_c = m\omega^{2}r\) and the weight \(F_g=mg\) act in the same direction (down - ward). The adhesive force \(F_a\) must counteract \(F_c + F_g\). At point II, the centripetal force is horizontal, and the adhesive force mainly counteracts the change in direction due to centripetal acceleration. At point III, the centripetal force is upward and the weight is downward. The adhesive force \(F_a\) needs to provide \(F_c - F_g\). At point IV, the centripetal force is horizontal.

Step2: Compare adhesive force requirements

Since \(F_c + F_g>F_c - F_g\) (assuming \(F_c> F_g\) as \(\omega\) is non - zero) and the horizontal centripetal force at II and IV requires less adhesive force compared to the vertical combination at I.

Step1: Apply Newton's second law

At point III, using Newton's second law \(F_{net}=ma\). The net force towards the center (centripetal force) \(F_c = m\omega^{2}r\). The forces acting on the ant are the adhesive force \(F_a\) (upward) and the weight \(mg\) (downward). So, \(F_a - mg=m\omega^{2}r\).

Step2: Solve for the adhesive force

Rearranging the equation \(F_a=m\omega^{2}r + mg\)

Step1: Recall centripetal acceleration formula

The centripetal acceleration \(a = \frac{v^{2}}{r}\), where \(v\) is the speed (constant) and \(r\) is the radius of the circular path. When the dancer is on the straight segments \(PQ\) and \(RS\), the motion is linear (constant velocity, so \(a = 0\)). When on the semicircular arcs \(QR\) and \(SP\), there is centripetal acceleration.

Step2: Analyze the acceleration - time graph

Since the length of \(PQ = QR=RS = SP\), the time spent on each segment is the same. The acceleration is non - zero (constant, because \(v\) and \(r\) are constant for the semicircles) only on the semicircular parts.

Answer:

A. I

Question 11