# Block sliding along a looped track

1. Jun 22, 2010

### LarryWineland

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
∑F=ma, 1/2 mv12 + mgy1 = 1/2mv22 + mgy2
Ac=v2/r

3. 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!!!

#### Attached Files:

• ###### Looptrack.jpg
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2. Jun 22, 2010

### 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!

3. Jun 23, 2010

### 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.