A small sphere of mass m = 40 grams is placed in a bucket and swung around in a

Question

A small sphere of mass m = 40 grams is placed in a bucket and swung around in a vertical circle with radius r = 30 cm (see figure below).

(a) What is the magnitude of the net force acting on the sphere when it is at position 1?

(b) What is the magnitude of the contact force on the sphere at position 1?

(c) At position 1, the contact force on the sphere is __________ times its weight.

(d) Now analyze the forces acting on the sphere when it finds itself at position 2. To begin with, calculate the magnitude of the contact force on the sphere at position 2.

(e) At position 2, the contact force on the sphere is ________ times its weight.

(f) Suppose you increase the swing speed from its previous value to 1.4 [m/s]. Calculate the new contact force on the sphere at position 2. ________ [N]

(g) this contact force controls whether the object maintains contact with the floor or not. If the contact force somehow is non-zero (as shown in class), then the object remains stuck to the floor even though the bucket is upside down. On the other hand, if this contact force becomes zero, then it is possible for the object to lose contact with the floor, even though the bucket is right side up.

This interplay between the centripetal force and the contact force is the principle behind the ferris wheel. You pay for the thrill of experiencing a situation in which the contact force on you has been carefully arranged to be equal to zero at a specific location on the ride.

Calculate the critical speed at which the contact force becomes zero. _________[m/s]

(h) At which location does the contact force become zero. Enter "1" for position 1, "2" for position 2 and "3" for neither.

(i) Calculate the new critical speed if I replace the sphere (m = 40 grams) with a much heavier object (m = 1 kg).

A small sphere of mass m = 40 grams is placed in a bucket and swung around in a vertical circle with radius r = 30 cm (see figure below). While doing this experiment, you find that it takes 20 seconds for the bucket to complete 10 revolutions. Answer the following questions: (a) What is the magnitude of the net force acting on the sphere when it is at position 1? (b) What is the magnitude of the contact force on the sphere at position 1? (c) At position 1, the contact force on the sphere is __________ times its weight. (d) Now analyze the forces acting on the sphere when it finds itself at position 2. To begin with, calculate the magnitude of the contact force on the sphere at position 2. (e) At position 2, the contact force on the sphere is ________ times its weight. (f) Suppose you increase the swing speed from its previous value to 1.4 [m/s]. Calculate the new contact force on the sphere at position 2. ________ [N] (g) this contact force controls whether the object maintains contact with the floor or not. If the contact force somehow is non-zero (as shown in class), then the object remains stuck to the floor even though the bucket is upside down. On the other hand, if this contact force becomes zero, then it is possible for the object to lose contact with the floor, even though the bucket is right side up. This interplay between the centripetal force and the contact force is the principle behind the ferris wheel. You pay for the thrill of experiencing a situation in which the contact force on you has been carefully arranged to be equal to zero at a specific location on the ride. Calculate the critical speed at which the contact force becomes zero. _________[m/s] (h) At which location does the contact force become zero. Enter 1 for position 1, 2 for position 2 and 3 for neither. (i) Calculate the new critical speed if I replace the sphere (m = 40 grams) with a much heavier object (m = 1 kg).

Explanation / Answer

Angular velocity = 10*2pi/20 =3.14rad/s => v=0.942 m/s

mass = 0.04kg R = 0.3m

a. Net force is zero as Contact force balances centrefugal force and weight.

b. Contact force = mv^2/R + mg = 0.51N

c. Contact force/weight = 1.04

d. Here contact force and centrefugal force bance weight. Contact force = mg - mv^2/R = 0.274N

e. Contact force/weight at position 2 = 0.698

f. New contact force = mg - m*1.4^2/R = 0.131N

g. Contact force = 0 when mg = mv^2/R. v=critical speed = 1.71m/s

h. Position 2

i. Critical speed does not depend on mass as it is = squareroot(r*g). So still 1.71m/s

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