What Is the Correct Method to Solve Rolling Motion Problems on an Incline?

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Homework Help Overview

The discussion revolves around rolling motion problems involving a wheel and a hoop on inclined planes, focusing on the application of conservation of energy and the calculation of moment of inertia. Participants are exploring the relationships between linear and angular motion as well as the implications of different assumptions about the objects' shapes and speeds.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of conservation of energy to relate potential energy to kinetic energy for both the wheel and the hoop. There are questions about the correct moment of inertia for the wheel and the distinction between final and average speeds in the context of the problems.

Discussion Status

Some participants have provided guidance on checking the moment of inertia and clarifying the need for final versus average speed in different scenarios. There is ongoing exploration of the calculations and assumptions made in the original poster's attempts, with no explicit consensus reached yet.

Contextual Notes

Participants are working within the constraints of typical homework problems, which may impose specific methods or assumptions about the objects involved, such as treating the wheel as a symmetric disk and the hoop as a standard hoop. There are indications of confusion regarding the definitions of final and average speeds in the context of rolling motion.

Soaring Crane
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Please check my setups and kindly tell me the correct method of finding the answers if possible.

1) The radius of a 3.0 kg wheel is 6.0 cm. The wheel is released from rest a point A on a 30 degree incline. The wheel rolls without slipping and moves 2.4 m to point B in 1.20 s. The moment of inertia of the wheel is closest to:

a.0.0057 kg*m^2
b.0.0048 kg*m^2
c.c. 0.0051 kg*m^2
d.0.0054 kg*m^2
e.0.0060 kg*m^2

At first, I thought it would be easy by just using I = 0.5*m*r^2 = 0.0054 kg*m/s^2.

Then, I thought of the conservation of energy.

mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

h = L*sin(30) = 1.2 m, where L = 2.4 m

Is v = 2.4 m/ 1.2 s = 2m/s? w = (2 m/s)/(.06 m) = 33.33 rad/s
I have all values so isolate I.
I = (2*29.28 J)/ (w^2) = (2*29.28 J)/ (33.33 rad/s)^2 = 0.0521 kg*m^2

This doesn’t match any of the choices.
What did I do wrong?


2) A hoop is released from rest at the top of a plane inclined at 24 degrees above horizontal. How long does it take the hoop to roll 24.0 m down the plane?

a. 3.27 s
b. 4.91 s
c. 4.48 s
d. 7.70 s
e. 3.13 s


I used conservation of energy for this part. I = M*r^2 for a hoop.

If L, length of incline, is 24 m then hoop must travel, height = L*sin (24) = 9.76168 m.

mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

mgh = 0.5*mv^2 + 0.5(m*r^2)*(v/r)^2

mgh = 0.5m*v^2 + 0.5*m*v^2

mgh = m*v^2

v = sqrt(g*h) ?

If I do that, v = sqrt(9.8 m/s^2*9.76168 m) = 9.78998 m/s

24 m/9.78998 m/s = 2.4514 s
What did I do incorrectly??

Thank you.
 
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You might need to check your moment of inertia. What does the wheel look like?
 
Well, from the side view in the diagram it is a circle (like O for cylinders) that is rolling down an incline.
 
prob #1

Soaring Crane said:
1) The radius of a 3.0 kg wheel is 6.0 cm. The wheel is released from rest a point A on a 30 degree incline. The wheel rolls without slipping and moves 2.4 m to point B in 1.20 s. The moment of inertia of the wheel is closest to:

a.0.0057 kg*m^2
b.0.0048 kg*m^2
c.c. 0.0051 kg*m^2
d.0.0054 kg*m^2
e.0.0060 kg*m^2

At first, I thought it would be easy by just using I = 0.5*m*r^2 = 0.0054 kg*m/s^2.
Looks like a match to me. But that assumes the wheel is a perfectly symmetric disk.

Then, I thought of the conservation of energy.

mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

h = L*sin(30) = 1.2 m, where L = 2.4 m

Is v = 2.4 m/ 1.2 s = 2m/s? w = (2 m/s)/(.06 m) = 33.33 rad/s
I have all values so isolate I.
I = (2*29.28 J)/ (w^2) = (2*29.28 J)/ (33.33 rad/s)^2 = 0.0521 kg*m^2

This doesn’t match any of the choices.
What did I do wrong?
That's the average speed, not the final speed at the bottom. (It starts from rest.)
 
prob #2

Soaring Crane said:
2) A hoop is released from rest at the top of a plane inclined at 24 degrees above horizontal. How long does it take the hoop to roll 24.0 m down the plane?

a. 3.27 s
b. 4.91 s
c. 4.48 s
d. 7.70 s
e. 3.13 s


I used conservation of energy for this part. I = M*r^2 for a hoop.

If L, length of incline, is 24 m then hoop must travel, height = L*sin (24) = 9.76168 m.

mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

mgh = 0.5*mv^2 + 0.5(m*r^2)*(v/r)^2

mgh = 0.5m*v^2 + 0.5*m*v^2

mgh = m*v^2

v = sqrt(g*h) ?

If I do that, v = sqrt(9.8 m/s^2*9.76168 m) = 9.78998 m/s

24 m/9.78998 m/s = 2.4514 s
What did I do incorrectly??

Thank you.
You made the same error as in prob #1, only in reverse. (The v you calculated is the final speed, not the average speed.)
 
For #1, I need to find the final speed, but, in #2, I need to find the average speed??:confused:
 
Soaring Crane said:
For #1, I need to find the final speed, but, in #2, I need to find the average speed??:confused:
Yes. In #1, you are finding the energy at the bottom--thus final speed is needed. In #2, you are using t = d/v, which is only true for constant speed or average speed.
 
How do I find the final speed for #1 and the average speed for #2?


For the final speed, do I have to use a kinematics equation?

For #2, the initial speed is 0 m/s and the final is 9.78998 m/s.
v^2 = 2*a*x
a = (9.78998)^2/(2*24 m) = 1.9967 m/s^2

v_f = a*t
t = (9.78998 m/s)/(1.9967 m/s^2) = 4.903 s for #2 ?
 
Last edited:
That's fine, but there's an easier way to find the average speed from the final speed. Since the acceleration is constant, V(ave) = (Vi + Vf)/2.
 
  • #10
For #1, I found:

the final velocity is v_f = 2*2m/s = 4 m/s if v_i = 0.

Now I have to change my kinetic engergy value.

0.5*(3 kg)*(4 m/s)^2 = 24 J

PE = 3Kg*9.8*1.3 m = 35.28 J

11.28 J = (I*w^2)/2

I= 2*11.28 J/(w^2)

w = 4 m/s/0.06 m = 66.67 rad/s

I = (2*11.28)/(66.67^2) = 0.005075 kg*m^2
 
  • #11
Looks OK to me.
 

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