# Rolling Objects on an Incline

1. Nov 17, 2006

### 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.

2. Nov 17, 2006

### physics girl phd

You might need to check your moment of inertia. What does the wheel look like?

3. Nov 17, 2006

### Soaring Crane

Well, from the side view in the diagram it is a circle (like O for cylinders) that is rolling down an incline.

4. Nov 17, 2006

### Staff: Mentor

prob #1

Looks like a match to me. But that assumes the wheel is a perfectly symmetric disk.

That's the average speed, not the final speed at the bottom. (It starts from rest.)

5. Nov 17, 2006

### Staff: Mentor

prob #2

You made the same error as in prob #1, only in reverse. (The v you calculated is the final speed, not the average speed.)

6. Nov 17, 2006

### Soaring Crane

For #1, I need to find the final speed, but, in #2, I need to find the average speed??

7. Nov 17, 2006

### Staff: Mentor

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.

8. Nov 17, 2006

### Soaring Crane

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: Nov 17, 2006
9. Nov 17, 2006

### Staff: Mentor

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. Nov 17, 2006

### Soaring Crane

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. Nov 17, 2006

### Staff: Mentor

Looks OK to me.