SUMMARY

Uniform circular motion, magnitude of velocity stays constant but direction changes. Since acceleration is defined as change in velocity over time taken, the change in direction (even though the magnitude stays the same) an acceleration is produced. At any point on the circle, the direction of the velocity is the tangent to the circle.

Centripetal acceleration a_c = v²/r

Centripetal force F_c = ma_c = mv²/r

The direction for both the centripetal acceleration and force is toward the center of the circle of motion. (NOTE: These two formulae will be used later. Every time a problem mentions circular motion remember these two formulae.)

Period T = 2pr/v (time to complete 1 revolution)

Newton's law of universal gravitation (This law says that there is an attractive force between any two objects and it is the force that keeps the planets going around the sun.) This force is one of the four fundamental forces in nature.

F = G m₁m₂/r²
where G is a constant whose value is 6.67´10^-11 N·m²/kg²,
m₁ and m₂ are the two masses, and r is the distance between the centers of the two masses.

Near earth surface

g = GM_E/r_E²
where M_E and r_E are the mass and radius of the earth.

For an object of mass M and the acceleration due to gravity at a distance r away is

g = GM/r²

Homework solutions

1. Jet plane traveling at speed of 1800 km/h = 500 m/s pulls on a radius of 5.0 km.

centripetal acceleration a_c = v²/r = (500 m/s)²/(5000 m) = 50 m/s².

In terms of "g", a_c = 50 m/s²/9.8 m/s² g = 5.1 g's.

8. Maximum speed a 1000-kg car can turn on radius 80 m on a flat road if m_s = 0.70 (between tires and road)? Is this result independent on the mass of the car?

The frictional force that keeps the car going round a bend

F_fr = m_sF_N where F_N = mg

= m_smg

This force is responsible for the centripetal acceleration.

i.e., F_fr = mv²/r

or m_smg = mv²/r




From the expression for v we note that it does not depend on the mass m. This implies that a small car or a large truck can corner at the same maximum speed provided of course they have the same coefficient of friction between the tires and the road.

10. Radius of rotation r = 10.0 m (in horizontal circle). Force felt by trainee is 7.75´ her own weight. Find how fast she is rotating.

If m is the mass of the trainee, then her weight is

W = mg

Given that the centripetal force is 7.75´ her weight

i.e., F_c = 7.75mg

This force comes from the rotation, i.e.,

F_c = 7.75mg = mv²/r, where v²/r is the centripetal acceleration.

v =

= 27.56 m/s.

To find the number of rev/s, we note that her speed at a radius of 10.0 m is 27.56 m/s.

Distance for 1 complete cycle is 2pr

revolution in 1 s is v/(2pr) = 27.56 m/s)/(2p´10.m) = 0.44 rev/s

(A simple formula for rev/s is

rev/s = 1/T = v/(2pr) where T is measured in seconds.)

15. Rotor-ride at a carnival.

Given rotation frequency f = 0.5 rev/s

Thus period T = 1/f = 2.0 s

Speed v = 2pr/T = 2p(5.0 m)/(2 s) = 15.7 m/s

Centripetal force F_c = mv²/r. This force is equal to F_N. (F_N means the force normal to the surface. The surface in this case is the wall so F_N is pointing horizontally toward the center of the rotor-ride.)

For a person not to slide down the vertical cylinder.

F_fr = mg

i.e., m_smv²/r = mg

m_s = gr/v² = (9.8 m/s²)(5.0 m)/(15.7 m/s)² = 0.20

The minimum coefficient of static friction is 0.20.


There is no outward force pressing the riders on the wall. It is an illusion. Since our natural tendency (Newton's first law of motion) without any net external force, is to go in a straight line, the rider's back is "hitting" the wall as the wall goes round in a circle. It is the "hitting" that gives the impression of an outward force.

23. Estimate the gravitational force between a man and a woman. Mass of woman m_w = 60 kg, mass of man m_m = 80 kg and distance r = 5.0 m.

Gravitation force

= 1.28´10^-8 N.

If r is 0.3 m,

= 3.56´10^-6 N.

These forces are very small. Clearly, it is not the force that attracts a man to a woman or vice versa!

28. Surface of a certain planet has gravitational acceleration g_p = 12.0 m/s². Brass of mass 3.0 kg on earth.

a. Mass of the brass on earth and on the planet.

Mass on earth m = 3.0 kg

Mass on planet m = 3.0 kg.

Note: Mass is independent on where it is placed.


b. Weight on earth and on the planet.

Weight on earth W_E = mg = (3.0 kg)(9.8 m/s²) = 29.4 N.

Weight on planet W_p = mg_p = (3.0 kg)(12.0 m/s²) = 36 N.

The weight is always the mass times the acceleration due to gravity at the point where the mass is located.

34. Acceleration on mars g_m = 0.38g_E where g_E is acceleration near earth's surface. Mars radius r_m = 3400 km. Find mass of mars.

The acceleration due to gravity near a planet's surface is given by

Solving gives M_mars = 6.45´10²³ kg.

39. Determine the time it takes a satellite to orbit the earth in a circular "near-earth" orbit.

Let m be the mass of the satellite near the surface. Then the force due to gravity is

F_g = mg where g = 9.8 m/s²

This force must be equal to the centripetal force, i.e.,

mg = mv²/r_E



The time it takes to complete one orbit is

From v = Ö(gr_E), we note that the velocity and thus the time T does not depend on the mass of the satellite.

41. What does a spring scale read for a 58.0 kg woman in an elevator that is:

a. moving upward with constant speed 6.0 m/s
b. moving downward with constant speed 6.0 m/s
c. moving upward with constant acceleration 0.33g
d. moving downward with constant acceleration 0.33g
e. freefall
Freebody diagram


a. Take up to be positive, SF = ma.

W - mg = ma = 0 since constant speed.

W = mg = (58 kg)(9.8 m/s²) = 568.4 N

Scale reads 58 kg.


b. Again W - mg = 0

Comparing with part (a), the scale reads 58 kg.

c. W - mg = ma = m(0.33g) since acceleration up is +

\ W = m(1 + 0.33) g

= 756 N

Scale reads 78 kg

d. W - mg = ma = m(-0.33g)

W = m(1 - 0.33)g

= 381 N

Scale reads 39 kg.

e. W - mg = ma

In free-fall a = -g

\ W = mg - mg = 0

i.e., the scale reads zero. This is a weightless condition.