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

• Soaring Crane
The difference between your method and mine is that I used the average speed instead of the final speed. Yours is more accurate. In summary, the moment of inertia for a 3.0 kg wheel released from rest on a 30 degree incline, rolling without slipping and moving 2.4 m in 1.20 s, is 0.005075 kg*m^2. In order to find this, the conservation of energy and the moment of inertia equations were used to calculate the final speed and average speed, respectively. For problem #1, the final speed was found to be 4 m/s, while for problem #2, the average speed was found to be 9.78998 m/s.
Soaring Crane
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.

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??

Soaring Crane said:
For #1, I need to find the final speed, but, in #2, I need to find the average speed??
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.

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

Looks OK to me.

## 1. What factors affect the speed of a rolling object on an incline?

The speed of a rolling object on an incline is affected by the angle of the incline, the mass and shape of the object, and the force of gravity.

## 2. How does the angle of the incline affect the motion of a rolling object?

A steeper incline will result in a faster speed for a rolling object, while a shallower incline will result in a slower speed. This is because a steeper incline provides a greater gravitational force, allowing the object to accelerate more quickly.

## 3. What is the difference between rolling and sliding motion on an incline?

When an object is rolling on an incline, it is rotating as it moves down the incline. This results in a combination of translational motion (movement in a straight line) and rotational motion. Sliding motion, on the other hand, involves only translational motion and does not involve rotation.

## 4. How does the mass of a rolling object affect its motion on an incline?

The mass of a rolling object affects its motion on an incline in two ways. First, a heavier object will experience a greater gravitational force, resulting in a higher speed. Second, a larger mass will also result in a greater moment of inertia, making it more difficult for the object to accelerate and maintain its speed.

## 5. What is the significance of the coefficient of rolling friction in the motion of a rolling object on an incline?

The coefficient of rolling friction is a measure of the resistance an object experiences while rolling. A lower coefficient of rolling friction will result in a smoother and faster motion, while a higher coefficient will result in a slower and more bumpy motion. This coefficient is important to consider when calculating the speed and acceleration of a rolling object on an incline.

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