Please check my setups and kindly tell me the correct method of finding the answers if possible.(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Rolling Objects on an Incline

**Physics Forums | Science Articles, Homework Help, Discussion**