171 Problem Set 12

Energy Conservation and Rotational Energy
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This problem set deals with energy conservation and rotational energy.
1.

The quantity in a rotational mechanical system that is like mass in a linear system is moment of inertia.
2. In the equation for rotational kinetic energy, Ek = ½Iw2, I is moment of inertia and w is angular speed.  
3.

If the formula for the moment of inertia of a 550 kg flywheel with a 0.3 m radius is given as I = ½mr2,

  1. what is the value of the moment of inertia?
  2. If it rotates at 600 rpm, what is its kinetic energy?
  
Knowns Unknowns Equations/Remarks
m = 550 kg
r = 0.3 m
w = 600 rpm
1 rev = 2p rad
a. I = ?
b. Ek = ?		 I = ½mr2 
Ek = ½Iw2 

a. I = ½mr2 = (1/2)(550 kg)(0.3 m)2
I = 24.75 kg·m2

b. Ek = ½Iw2 
Ek = (1/2)(24.75 kg·m2)[(600 rev/min)(2p rad/1 rev)(1 min/60 sec)]2
Ek = 48,900 J

Substitute in the known values and calculate the answer.

Convert rpm to rad/s. We end up with units of kg·m2/s2, which is joules.

4. How much kinetic energy can be stored in a 5,000 lb flywheel with a 3 ft radius that rotates at 9,000 rpm?

  
Knowns Unknowns Equations/Remarks
w = 5000 lb
r = 3 ft
w = 9000 rpm
g = 32 lb/slug (makes for easier unit cancellation)
1 rev = 2p rad 		Ek = ? w = mg
I = ½mr2 
Ek = ½Iw2 

w = mg
m = w/g

I = ½mr2 = (1/2)(w/g)r2
I = (1/2)(5000 lb/32 lb/slug)(3 ft)2
I = 703.1 slug·ft2

Ek = ½Iw2 
Ek = (1/2)(703.1 slug·ft2)[(9000 rev/min)(2p rad/1 rev)(1 min/60 sec)]2
Ek = 3.12 ×108 ft·lb

Rearrange the weight equation to solve for mass, then substitute in the moment of inertia equation. Using 32 lb/slug makes canceling units easier (basically 1 slug weighs 32 lbs).

Substitute the known values in the kinetic energy equation. Convert the rpm to rad/s. A pound is slug·ft/s2, so the units end up in ft·lbs.

5.

For the energy systems we have been studying, any energy not accounted for as potential energy or kinetic energy is lost energy. For conservation of energy to occur, what kind of energy is the lost energy most likely to be?

Heat energy

  
6. A block with a mass of 10 kg is located 30 meters above a road. If the block fell, what would be its kinetic energy at the moment it struck the road?

  
Knowns Unknowns Equations/Remarks
m = 10 kg
h = 30 m
g = 9.8 N/kg
Ek at 0 m = Ep at 30 m 		Ek = ? Ep = mgh 

Ek = Ep = mgh 

Ek = (10 kg)(9.8 N/kg)(30 m)

Ek = 2940 J

The potential energy of the block at 30 m is converted into kinetic energy. At 0 m the block's kinetic energy is equal to it's original potential energy (assuming no losses to air resistance). Substitute in the known values and calculate the answer.
7.

A system has a total energy of 280 ft·lb. If there are no losses of heat energy and 100 ft·lb is potential energy, describe the remaining energy in the system.

The remaining energy is kinetic energy and has the value of 180 ft·lb, from the equation Etotal = Ep + Ek + Losses (Losses = 0).

  
8.

A sphere has a mass of 8 kg and a radius of 0.13 m and rotates at 400 rpm.

  1. Find the moment of inertia.
  2. Find the rotational kinetic energy.
  
Knowns Unknowns Equations/Remarks
m = 8 kg
r = 0.13 m
w = 400 rpm
1 rev = 2p rad 		a. I = ?
b. Ek = ?		 I = 2/5mr2 
Ek = ½Iw2 

a. I = 2/5mr2 = (2/5)(8 kg)(0.13 m)2
    I = 0.0541 kg·m2

b. Ek = ½Iw2 
    Ek = (1/2)(0.0541 kg·m2)[(400 rev/min)(2p rad/1 rev)(1 min/60 sec)]2
    Ek = 47.5 J

Substitute the known values in the equations and calculate the answers. Remember to convert rpm to rad/s.
9. A 12 V battery provides a current of 0.2 A. How much energy is expended in 40 seconds?

  
Knowns Unknowns Equations/Remarks
V = 12 V
I = 0.2 A
t = 40 sec		 E = ? E = VIt 

E = VIt =(12 V)(0.2 A)(40 sec)

E = 96 J

A volt is J/C and an ampere is C/s. The units end up being (J/C)(C/s)(s) joules (J).
10.

A 40 kg flywheel in the shape of a solid disc has a radius of 8.0 cm and is rotating at 900 rpm.

  1. Find the kinetic energy stored in the flywheel.
  2. The flywheel is used to turn an electrical generator, but 70% of its energy is lost due to friction. How much electrical energy will be produced when the flywheel comes to a complete stop?
  3. The average voltage produced by the generator is 12 V over a period of 2.0 minutes. What will the average current be?
  
Knowns Unknowns Equations/Remarks
m = 40 kg
r = 8 cm = 0.08 m
w = 900 rpm
V = 12 V
t = 2.0 min = 120 sec
1 rev = 2p rad
1 m = 100 cm
1 min = 60 sec

a. Ek = ?
b.Eelect = ?
c. I = ?

 Ek = ½Iw2 
I = ½mr2 
Eelect = 0.3(Ek) 
E = VIt 

a. I = ½mr2 = (1/2)(40 kg)(0.08 m)2
    I = 0.128 kg·m2

    Ek = ½Iw2 
    Ek = (1/2)(0.128 kg·m2)[(900 rev/min)(2p rad/1 rev)(1 min/60 sec)]2
    Ek = 568.5 J

b. Eelect = 0.3(Ek) = 0.3(568.5 J)
    Eelect = 170.6 J

c. E = VIt 
    I = E/(Vt) = (170.6 J)/[(12 V)(120 sec)]
    I = 0.118 A or 118 mA

Find the moment of inertia for the flywheel. Use this to find the kinetic energy. Convert rpm to rad/s.

Seventy percent of the energy was lost, so only 30% remains.

The electrical units work out to be amperes (C/s). Refer to problem 9's remarks.
11.

A heavy spinning flywheel has a radius of 2.0 meter. It requires 600 J of energy to bring it to a stop in one revolution.

  1. Find the braking torque applied to the flywheel.
  2. Find the braking force required to stop the flywheel in one revolution.
  
Knowns Unknowns Equations/Remarks
r = 2.0 m
W = 600 J
q = 1 rev 		a. t = ?
b. F = ?		 W = t·q 
t = F × r (r is the lever arm, or radius in this case)

a. W = t·q 
    t = W/q = (600 J)/[(1 rev)(2p rad/1 rev)]
    t = 95.5 N·m

b. t = F × r
    F = t/r = (95.49 N·m)/(2.0 m)
    F = 47.7 N

Rearrange the work equation to find the torque needed to stop the flywheel in one revolution.

Rearrange the equation to find the force needed to create the required torque.

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