Rotational Motion and Conservation of Energy

[AikiGirl]

Homework Statement

A uniform solid sphere rolls on a horizontal surface at 20m.s^-1 and then rolls up an incline which has an angle on inclination of 30°. Ignoring friction, calculate the height attained by the sphere.

Homework Equations

Weren't given any specific equations to work with but here are some I've attempted to use:
Ek(lin)= 1/2 mv^2, Ek(rot)=1/2 Iw^2, Ep=mgh, Vf=Vi+at, D=Vi t + 1/2 at^2

The Attempt at a Solution

As it is a uniform sphere I ignored Ek(rot). We were given an answer of 29m.

E(before)=E(after)
Ek+Ep=Ek+Ep
1/2mv^2 +mgh= 1/2mv^2 +mgh (masses cancel)
0.5x20^2 + 9.8x0 = 0.5 (20/cos30)^2 + 9.8h Not sure if this is the right one to use
200 = 266.67 + 9.8h
-66.67 = 9.8h
h= - 6.8

So I tried looking at just the vertical component:
Vf=Vi+at
0=(20tan30) + 9.8 t
9.8t=11.55
t=1.18s

D=Vi t + 1/2 at^2
D=11.55 x 1.18 + 0.5 x 9.8 x 1.18^2
D= 20.41m

Does anyone see where I went wrong? Any help would be greatly appreciated!

 

[ehild]

The Attempt at a Solution

As it is a uniform sphere I ignored Ek(rot). We were given an answer of 29m.

Uniformity does not mean ignorable rotational energy.

ehild

 

[AikiGirl]

I still can't get the right answer?

Ek(rot) + Ek(lin) +Ep = Ek(rot) + Ek(lin) +Ep
1/2 Iw^2 + 1/2mv^2 +mgh = 1/2 Iw^2 + 1/2mv^2 +mgh
1/2 r^2(v^2)/(r^2) + 1/2 v^2 +0 = 0 + 0 + 9.8h
1/2 x 20^2 + 1/2 x 20^2 =9.8h
200+200 = 9.8h

h=40.8m

Is that right working??

 

[da_nang]

I still can't get the right answer?

Ek(rot) + Ek(lin) +Ep = Ek(rot) + Ek(lin) +Ep
1/2 Iw^2 + 1/2mv^2 +mgh = 1/2 Iw^2 + 1/2mv^2 +mgh
1/2 r^2(v^2)/(r^2) + 1/2 v^2 +0 = 0 + 0 + 9.8h
1/2 x 20^2 + 1/2 x 20^2 =9.8h
200+200 = 9.8h

h=40.8m

Is that right working??

Not quite. What's the moment of inertia for a uniform solid sphere?

 

[AikiGirl]

According to wiki it's I = (2mr^2)/5

So,

1/2 x 2/5 x 20^2 + 200 =9.8h
280=9.8h
h=28.57m

:D