A ball at the end of a 2-m rope swings in a vertical arc. Its speed at the bottom of the swing is 8 m/s. What will be its speed when it is at a position where the rope is horizontal?

Question

A ball at the end of a 2-m rope swings in a vertical arc.  Its speed at the bottom of the swing is 8 m/s.  What will be its speed when it is at a position where the rope is horizontal?

anonymous · Answer

v= square root of (m*length*(1-costheta) )  and theta is 90 right?? :( where am i going wrong?

jamesj · Answer

Use conservation of energy.  At the bottom all the energy is kinetic
$$ KE = \frac{1}{2}mv^2 = 32m $$
while when the rope is horizontal it has a mixture of KE' (let's call it KE' to differentiate it from the other KE) and PE.  The PE is
$$ PE = mgh = 19.6m $$ as h = 2
Now find the KE' at this point and solve for v' in that KE'

jamesj · Answer

I.e.,
KE = PE + KE'

You now have expressions for KE and PE, use those to write down an expression for KE' and solve for v.

anonymous · Answer

I dont understand what your using for mass... ??? 1kg? or 0? If it was 0 then that couldnt be right...

jamesj · Answer

I didn't put any mass in.  I just left that as a variable and that is why 'm' appears in both of the expressions for KE and PE.  

Keep on going and you'll see you can cancel it out.

anonymous · Answer

ok we thought it was meters.... i was confused...okokok one second

anonymous · Answer

-5N

anonymous · Answer

5 m/s sorry

jamesj · Answer

right

anonymous · Answer

velocity at horizontal position=root of(square of(initial velocity)-(2gr));
                                          =root of((8*8)-(2*10*2))
                                          =root of(64-40)=root of 24=5 m/s(about)