Physics Chapter 7 Rotational Motion and Gravitation

Rotational motion has to do with gears, wheels, tools, orbits, and roller coasters. This is how tires roll and how amusement rides work. This is how meteors buzz past our planet and how we could one day reach mars.

We are going to have to learn new variables.

Why does the arc length formula θ = S/R include θ and not tan θ like  regular trigonometry? - Mathematics Stack Exchange
Theta we all know as the angle. r we know as the radius of the circle and now we are adding s for the arc length. s / r = theta. In this section we are going to be using theta in terms of radins instead of degree. 360 degrees will equal 2(pi) radians. 2(pi) radians will equal 1 revolution. 1 radian is about 57.3 degrees.

It may be a good idea to brush up on the conversion of degrees and radians

From linear motion, we have basic variables for displacement, velocity, and acceleration. In rotational motion, we will also have similar variables they will just have new symbols.

These variables have not only relationships with the same type of motion but also relationships with linear motion.



Let’s do some homework!

  1. Convert 45.0 degrees to radians, 11.0 radians to revolutions, and 78.0 rpm into rad/s.

To convert degrees to radians we need to multiply the 46 degrees by (pi * rad)/180 degrees. the degrees will cancel out and leave the rads.

To convert radians to revolutions we need to multiply 11 radians by (1 revolution)/ (2 * pi * rad). The radians will cancel out and leave the revolutions.

To convert rpm to revolutions we need to multiply 78 rpm by ( 1 minute / 60 seconds) and (2 * pi * rad / 1 revolution). The minutes and revolutions will cancel out and leave us with radians per second.

2. A potter’s wheel moves uniformly from rest to an angular speed of 0.24 rev/s in 26.0 s. (a) Find the angular acceleration in radians per second per second. (b) would doubling the angular acceleration during the given period have doubled final angular speed?

You will want to start out by getting revs per second per second. So we take the 0.24 rev/s and divide by the 26 s. Then we will multiple to convert to rads.

Yes, from rest the angular speed is related to the angular acceleration by the equation below, so by doubling the angular acceleration you will also be doubling the angular speed.

3. A dentist’s drill starts from rest. After 3.80 s of constant angular acceleration it turns at a rate of 2.80 x 104 rev/min. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

Using the same formula from before we can find the angular acceleration and then convert it into rad/s2

To find the angle we will use the following formula and convert to find radians

4. A bicyclist starting at rest produces a constant angular acceleration of 1.50 rad/s2 for wheels that are 33.5 cm in radius. (a) What is the bicycle’s linear acceleration (in m/s2)? (b) What is the angular speed of the wheels (in rad/s) when the bicyclist reaches 10.2 m/s? (c) How many radians have the wheels turned through in that time? (d) How far (in m) has the bicycle traveled?

From the problem we know that r = 33.5 cm which converts to 0.335 m (since we work in meters) and the angular acceleration [a] = 1.50 rad/s2 . [a] will stand for alpha.

For (a) we are looking for linear acceleration at which we know at = r*[a] so to find at substitute the values for r and [a]

For (b) we will start with formula vt = r vvf and rewrite it to solve for the linear speed

For (c) we are going to rewrite the time-independent rotational equation to find the angular displacement

For (d) to find distance you will need multiply r by the angular displacement

5. It has been suggested that rotating cylinders about 13.5 mi long and 4.02 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?

First to find r it is half of the diameter. so that will be 2.01 mi. Now convert miles to meters. (2.01 mi )( 1609 m / 1 mi ) = 3.23 x 103 m. Then using the acceleration of centripetal we need to find angular velocity.

Substrituting in the values that we know r is 3.23 x 103 m and ac is 9.80 m/s2 due to free fall

6. An adventurous archeologist (m = 82.0 kg) tries to cross a river by swinging from a vine. The vine is 11.5 m long, and his speed at the bottom of the swing is 9.00 m/s. The archeologist doesn’t know that the vine has a breaking strength of 1,000 N. Does he make it safely across the river without falling in?

7. Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person wearing a g-suit can withstand accelerations up to about 9g (88.2 m/s2) without losing consciousness. (a) If a human centrifuge has a radius of 3.81 m what is the angular speed results in a centripetal acceleration of 9g? (b) what linear speed would a person in the centrifuge have at this acceleration?

(a) we need to use the centripetal acceleration formula to solve for the angular angular velocity. Like in question 5. However, we are given the 9 g is equal to 88.2 m/s2. So substitute out values and we can find the angular velocity

(b) we are looking for linear velocity so it is a simple as multiplying the radius by the angular velocity

8. A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 28.5 m/s. With what maximum speed can it go around a curve having a radius of 61.5 m?

Using the friction between the tires and the roadway we can find the max centripetal acceleration first and then find the change in max speed


Then using that we can find the new max velocity


9. An air puck of mass m1 = 0.25 kg is tied to a string and allowed to revolve in a circle of radius R = 0.9 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of m2 = 1.5 kg is tied to it (see the figure below). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) what is the tension in the string? (b) what is the horizontal force acting on the puck? (c) what is the speed of the puck?

(a) we need to find the tension of the string and luck for us it will also be equal (b) to the horizontal force acting on the puck
(c) Now knowing the relation of since we know that centripetal force is related to linear velocity in the formula below. We can rewrite it to find the linear velocity.

10. A roller-coaster vehicle has a mass of 440 kg when fully loaded with passengers (see figure). (a) If the vehicle has a speed of 17.0 m/s at point A, what is the force (in N) of the track on the vehicle at this point? (b) What is the maximum speed (in m/s) the vehicle can have at point B in order for gravity to hold it on the track?

(a) For A we need to find the sum of the centripetal forces and normal force acting upon the car.

(b) the key for this is when the car is on B point it is about to leave the track so we set n = 0 and then solve for Vt

11. You are swinging a block of mass 0.323 kg tied to the end of a string of length 0.44 meters, in a vertical circle. You swing the block so that it has a constant linear speed(tangential) speed of 5.3 m/s. (a) what is the weight of the block? (b) what is the centripetal acceleration of the block? (c) what is the amount of tension in the string when the block is at the top of the circle? (d) what is the amount of tension in the string when the block is at the side of the circle? (e) what is the amount of tension in the string when the block is at the bottom of the circle?

12. Two objects attract each other with a gravitational force of magnitude 9.00  10-9 N when separated by 20.0 cm. If the total mass of the objects is 5.00 kg, what is the mass of each?

Firstly, we are going to need to know what G stands for. G = 6.67 x 10-11 Nm2 / kg2 this is the gravity constant. We are given Fg as 9.00 x 10-9 N and r = 20 cm which converts to 0.20 m. We also know the sum of the mass = 5. We need to rewrite the equation into a form that has the masses by themselves and go ahead and calculate that side of the equation. Then we can substitute mass 2 as [5.00 kg – mass 1]. This will allow us to use a quadratic equation to find the two values of mass.


13. A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circular, determine the mass of Mars.

Using Kepler’s 3rd law formulae we can get mass by itself with a little rework. Orbital radius will be Ms. Period will be T and G will be the constant.

14. A synchronous satellite, which always remains above the same point on a planet’s equator, is put in a circular orbit around Mars so that scientists can study a surface feature. Mars rotates once every 24.6 h.

15. In a popular amusement park ride, a rotating cylinder of radius R = 4.00 m is set in rotation at an angular speed of 5.90 rad/s, as in the figure shown below. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider’s clothing and the wall is needed to keep the rider from slipping? Hint: Recall that the magnitude of the maximum force of static friction is equal to 𝜇sn, where n is the normal force—in this case, the force causing the centripetal acceleration.

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