Can Mechanical Energy Be Conserved in an Accelerated Frame?

[A13235378]

Homework Statement

An observer located in the cart that accelerates with 7.5m / s ^ 2, realizes that when the 4kg block passes through the lowest part of the 20 cm radius circumference, it has a speed of 4m / s. Determine the magnitude of the force that the block exerts for such an observer at the highest part of the circumference (g = 10m / s ^ 2).

Relevant Equations

k = mv^2/2
F = m.a
P (potencial energy) = mgh

In the frame of the accelerated block, I applied the non-inertial force F'. My doubt is if I can conserve the mechanical energy in the accelerated frame and find the speed at the top. Otherwise, how could you proceed.

 

[etotheipi]

Energy conservation in a non-inertial frame works almost exactly like an inertial frame, so long as you include the works done by inertial forces.

However, note that in the non-inertial frame, the inertial force in this case is constant, and has no component in the vertical direction. So the work done by this non-inertial force is simply $-ma \mathbf{e}_x \cdot 2R \mathbf{e}_y = 0$, so you can essentially ignore it when writing down your energy equation between the bottom and top of the circle.

 

[A13235378]

Energy conservation in a non-inertial frame works almost exactly like an inertial frame, so long as you include the works done by inertial forces.

However, note that in the non-inertial frame, the inertial force in this case is constant, and has no component in the vertical direction. So the work done by this non-inertial force is simply $-ma \mathbf{e}_x \cdot 2R \mathbf{e}_y = 0$, so you can essentially ignore it when writing down your energy equation between the bottom and top of the circle.

First, thanks! Second, I conserved the energy:

mv^2/2 = mgh+ mv'^2/2
8 = 4 + v'^2/2
v'^2 = 8

So, I applied the resulting centripetal (I don't know if I could)

N + P = mv'^2/R
N + 40 = 160
N=120

Then, I made Pythagoras with the fictitious force:

F^2 = 120^2 + 30^2
F = 123,7 N

The answer is 75 . Where I am wrong?

 

[PeroK]

The answer is 75 . Where I am wrong?

Does the answer have a formula, or just $75N$?

 

[A13235378]

Does the answer have a formula, or just $75N$?

Just 75 N

 

[etotheipi]

If I'm not mistaken, the force the block exerts on the cart should just be the normal force. Remember that the inertial force only acts on the block, and isn't part of a force pair.

I think the answer is wrong, but better wait for someone else to confirm this!

 

[PeroK]

Just 75 N

I don't see how they get that.

 

[PeroK]

If I'm not mistaken, the force the block exerts on the cart should just be the normal force. Remember that the inertial force only acts on the block, and isn't part of a force pair.

I think the answer is wrong, but better wait for someone else to confirm this!

If we ignore gravity, then the inertial force acts like a fictitious gravity and the block will be traveling at $4m/s$ when it reaches it reaches the top - that's just sliding round a half-circle to the same as the starting "height". Therefore, as you previously suggested, the horizontal force does no net work. And the problem reduces to the same as one without the horizontal acceleration.

I get $120N$.

 

[etotheipi]

If we ignore gravity, then the inertial force acts like a fictitious gravity and the block will be traveling at $4m/s$ when it reaches it reaches the top - that's just sliding round a half-circle to the same as the starting "height". Therefore, as you previously suggested, the horizontal force does no net work. And the problem reduces to the same as one without the horizontal acceleration.

I get $120N$.

Yeah, same. And if PeroK also gets a different answer to the mark scheme, then we can be damn well sure that the mark scheme is incorrect

 

[PeroK]

Yeah, same. And if PeroK also gets a different answer to the mark scheme, then we can be damn well sure that the mark scheme is incorrect

I don't know about that!

 

[A13235378]

Thanks everbody

 

[PeroK]

First, thanks! Second, I conserved the energy:

mv^2/2 = mgh+ mv'^2/2
8 = 4 + v'^2/2
v'^2 = 8

So, I applied the resulting centripetal (I don't know if I could)

N + P = mv'^2/R
N + 40 = 160
N=120

Then, I made Pythagoras with the fictitious force:

F^2 = 120^2 + 30^2
F = 123,7 N

The answer is 75 . Where I am wrong?

I agree with this, except the inertial force is tangential to the drum at the top, so the block exerts only a vertical force equal to the centripetal force exerted by the drum on the block. I.e. $120N$, as you calculated.

 

[etotheipi]

And at an arbitrary angle $\theta$ from the downward vertical, $N - mg\cos{\theta} - ma\sin{\theta} = mv^2/R$ implies that$$N(\theta) = mg(3\cos{\theta} - 2) + 3ma\sin{\theta} + mv_1^2/R$$which does give $N(\pi) = 120 \text{N}$