Solving for Height of a Falling Cube Using Kinetic Energy

[goonking]

Homework Statement

Homework Equations

1/2 mv^2 = mgh

The Attempt at a Solution

first, a side question, if p = mv, can't we just place p into 1/2 mv^2 so it becomes 1/2 p^2?

anyway, on to the question, at the point where the cube falls off, the Fnormal should just about equal 0, correct?
is this the way to approach the problem? Or we can find the velocity where the cube loses contact with the sphere and plug that v into 1/2 mv^2 = mgh, solving for h.

 

[Pierce610]

3. No you can't because there is only v square while m isn't.
Personally I would prefer to consider the balance of two force, gravity and centripetal, looking at which speed the mass can still move of a rotational motion.

 

[goonking]

3. No you can't because there is only v square while m isn't.
Personally I would prefer to consider the balance of two force, gravity and centripetal, looking at which speed the mass can still move of a rotational motion.

so at the point the cube loses contact with the half-sphere, acceleration becomes 9.8 m/s^2.

a = v^2/r

9.8 x 10.3m = v^2
v= 10.04 m/s

1/2 m v^2 = mgh
mass cancels out

1/2 (10.04)^2 = g h

h = 5.15 meters.

is that correct?

 

[Satvik Pandey]

so at the point the cube loses contact with the half-sphere, acceleration becomes 9.8 m/s^2.

a = v^2/r

9.8 x 10.3m = v^2
v= 10.04 m/s

1/2 m v^2 = mgh
mass cancels out

1/2 (10.04)^2 = g h

h = 5.15 meters.

is that correct?

When the ball will freely fall then it's total acceleration will be $g$. You just need here radial acceleration.

If the block looses contact with the sphere then what can you say about the magnitude of Normal reaction force acting on the block.
If the block has velocity at $v$ at the instant shown then what should be the centripetal force acting on it? This centripetal force it provided by which force? Equate them and get the answer.

 

[goonking]

When the ball will freely fall then it's total acceleration will be $g$. You just need here radial acceleration.

View attachment 80691

If the block looses contact with the sphere then what can you say about the magnitude of Normal reaction force acting on the block.
If the block has velocity at $v$ at the instant shown then what should be the centripetal force acting on it? This centripetal force it provided by which force? Equate them and get the answer.

Centripetal force is provided by Fnormal? which is 0 when it loses contact with the half-sphere.

correct?

 

[Satvik Pandey]

Centripetal force is provided by Fnormal? which is 0 when it loses contact with the half-sphere.

correct?

Yes Normal force will be zero when it looses contact. But centripetal force is provided by the component of mg. Can you find it?

 

[goonking]

Yes Normal force will be zero when it looses contact. But centripetal force is provided by the component of mg. Can you find it?

Centri. Force = m a

a = v^2/r = mg ?

 

[Satvik Pandey]

Centri. Force = m a

a = v^2/r = mg ?

I said a $component$ of mg. Can you find it in terms of mg and $\theta$?

 

[goonking]

it should be the hypotenuse in the picture, which is mg / cos theta

how do we solve for theta?

 

[Satvik Pandey]

it should be the hypotenuse in the picture, which is mg / cos theta

how do we solve for theta?

It will be $mgcos\theta$

First equate this to centripetal force. You can find relation between $v$ and $\theta$ by conservation of energy.

 

[goonking]

It will be $mgcos\theta$

First equate this to centripetal force. You can find relation between $v$ and $\theta$ by conservation of energy.

mgcosθ = m v^2 / rgcosθ = v^2/r

correct?

 

[Satvik Pandey]

mgcosθ = m v^2 / rgcosθ = v^2/r

correct?

Yes. Now use conservation of energy to find relation between v and theta.

 

[goonking]

Yes. Now use conservation of energy to find relation between v and theta.

conservation of energy = 1/2 mvi^2 + mghi = 1/2 mvf^2 + mghf ?

 

[Satvik Pandey]

conservation of energy = 1/2 mvi^2 + mghi = 1/2 mvf^2 + mghf ?

What is the initial velocity of the block?
Can you represent hi and hf in terms of R and theta?

 

[goonking]

What is the initial velocity of the block?
Can you represent hi and hf in terms of R and theta?

initial velocity should be 0 since it was at rest.

hi is 10.3m

hf is what we are looking for

and i have no idea how to find theta.

 

[goonking]

.

 

[Satvik Pandey]

initial velocity should be 0 since it was at rest.

hi is 10.3m

hf is what we are looking for

and i have no idea how to find theta.

hf is $Rcos\theta$. Now proceed.

 

[goonking]

hf is $Rcos\theta$. Now proceed.

but we have 2 unknowns, hf and theta.

 

[Satvik Pandey]

but we have 2 unknowns, hf and theta.

$h_{f}=Rcos\theta$. Put this in energy equation. You would have two unknowns v and $\theta$. Remember the equation which you made by using centripetal force. Use these two equation and find $\theta$.

 

[goonking]

$h_{f}=Rcos\theta$. Put this in energy equation. You would have two unknowns v and $\theta$. Remember the equation which you made by using centripetal force. Use these two equation and find $\theta$.

is v = 14 m/s?

 

[Satvik Pandey]

is v = 14 m/s?

I think you have done some mistake. The equations are
$2mgR(1-cos\theta)=mv^{2}$

and $Rgcos\theta=v^{2}$

Find $\theta$. You don't have to find v.

 

[goonking]

I think you have done some mistake. The equations are
$2mgR(1-cos\theta)=mv^{2}$

and $Rgcos\theta=v^{2}$

Find $\theta$. You don't have to find v.

angle should be 48 degrees

 

[Satvik Pandey]

angle should be 48 degrees

Yes $cos\theta=\frac{2}{3}$. From this can you find the height above which it looses contact with the ball?

 

[goonking]

Yes $cos\theta=\frac{2}{3}$. From this can you find the height above which it looses contact with the ball?

6.89 m

thank you!

 

[Satvik Pandey]

6.89 m

thank you!

You are welcome!