Block sliding along a looped track

[LarryWineland]

Homework Statement

A block of mass m slides without friction along the looped track shown in figure 6-39. If the block is to remain on the track, even at the top of the circle (whose radius is r), from what minimum height h must it be released. Next, if the actual release height is 2h, calculate (b), the normal force exerted by the track at the bottom of the loop, (c) the normal force exerted by the track at the top of the loop, and (d) the normal force exerted by the track after the block exits the loop onto the flat section. SEE ATTACHED PICTURE FOR TRACK DIAGRAM.

Homework Equations

∑F=ma, 1/2 mv₁² + mgy₁ = 1/2mv₂² + mgy₂
A_c=v²/r

The Attempt at a Solution

I was able to get the first part (determining the minimum height) this way:
Assuming that the normal force on the block at the top of the track must be close to zero, this force would be the minimum necessary for the block to stay on the track. Both the normal force and weight (mg) would be pointing downward, and since the block is going in a circle (the loop), we must incorporate centripetal acceleration into Newton's 2nd law, and solve for v (you need to solve for v so that you can substitute it into the conservation of energy equation in order to find the minimum height).

I solved for v like this: ∑F= m(v²/r) ---> Fn + mg = m(v²/r)--- masses cancel---> 0+g=v²/r ---> gr=v²---> v=$$\sqrt{}gr$$

I then substituted $$\sqrt{}gr$$ for v into 1/2 mv₁²+mgy₁ =1/2mv₂²+mgy₂, and I chose y1 to be the height, and y2 to be the top of the loop, which would be 2r since the height of a circle is 2r. Solving for y1, I got h = 2.5r, which is correct.

However I haven't the slightest clue how to set up the variables for other parts of the problem. I even found the solution in the text and the explanation still makes no sense to me (the text is Giancoli 6th edition, problem is #40 and 75 from chapter 6). Any help would be greatly appreciated! Thanks!

 

[quantum13]

first draw a free body diagram. the only forces acting on it are gravity and the normal force. you also know the net acceleration because you know the velocity of the block at any height from energy considerations

sounds like a + x = b to me!

 

[inky]

At the top,direction of normal force and weight are towards the centre and at the bottom, opposite direction. velocity depends on the point for circular track. weight is same as any point but normal force is not the same. At the top normal force value is very small and at the bottom, it's value is greater.

(b)Use law of conservation of energy for height 2h and at the bottom. Find the velocity at the bottom.Then use Newton's 2nd law at the bottom for circular track. You may calculate normal force.
(c) same as (b)'s calculation but answers are not the same.

For (d) normal force is biggest value. It has no radial force. Consider normal force and weight.

Try it. We can help you.

 

[LarryWineland]

Thanks everybody. I figured it out with your help. It was simpler than I realized.

 

[rwisz]

I would like to commend you for your attempt at solving this problem. It is clear that you have a good understanding of the basic principles involved and have made a good start on the problem.

To solve the other parts of the problem, we can use the same approach of applying Newton's second law and conservation of energy. Let's start with part (b), where we need to find the normal force at the bottom of the loop.

At the bottom of the loop, the block is moving in a circular motion, so there must be a centripetal force acting on it. This force is provided by the normal force from the track. We can set up the equation ∑F=mv2/r and solve for the normal force (Fn). This will give us the normal force at the bottom of the loop.

For part (c), we need to find the normal force at the top of the loop. Here, we can use the same approach as in part (a) and set the normal force equal to zero. However, we need to remember that the block is now moving in a vertical circle, so the centripetal force is provided by the block's weight and the normal force. We can set up the equation ∑F=mv2/r and solve for the normal force (Fn). This will give us the normal force at the top of the loop.

Finally, for part (d), we need to find the normal force after the block exits the loop and goes onto the flat section. Here, we can use the conservation of energy equation and set the initial and final energies equal to each other. The initial energy will be the kinetic energy of the block at the top of the loop, and the final energy will be the kinetic energy of the block at the bottom of the flat section. We can solve for the normal force using the equation ∑F=mv2/r again.

I hope this explanation helps you understand the problem better. Keep up the good work!