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Cart + loop

I'm having trouble getting started on this entire problem. I've re-drawn it, I'm given the mass of the cart, and the radius of the loop.. but I have absolutely no idea how I can calculate even the first part of this (minimum high h).

I'd greatly appreciate any help anyone could offer. Thank you!!

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The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m, as shown in the figures. Assume that friction can be ignored. Also assume that, in order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.6 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?

b) For this part, we launch the cart horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" (as defined above) negotiate the loop?

c) When the car is descending vertically (ie at a height R above the ground) in the loop, what is its speed |v|?

d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 20 m. What average retarding acceleration |a| is required?
 

Attachments

ehild

Homework Helper
15,360
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Aki Yamaguchi said:
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The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 15 m, as shown in the figures. Assume that friction can be ignored. Also assume that, in order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.6 times the weight of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.) You may treat the cart as a point particle.

a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?
Let's start with this part. What do we know?
a. There is no friction. The mechanical energy is conserved. If y the cart is at the top of the loop,

[tex]0.5 mv^2=mgh[/tex]

b. In the loop, the cart performs circular motion. The centripetal force, needed to this motion with speed v results as the sum of both the normal (to the loop) component of gravity and the normal force exerted by the loop. At the top both point vertically downward.

[tex]Fcp=mg+Fn=mv^2/R[/tex]

c. For safety,

[tex]Fn\ge 0.6 mg[/tex]

Continue...

ehild
 

mukundpa

Homework Helper
524
3
I think that you have problem with minimum velocity required at the lowest point of the loop for safe negotiation of the loop. Actually if the velocity is not sufficient, the cart will leave the loop. so it must go to the heightest point and till then the normal reaction should not be zero. At heighest point the speed should be such that its weight provides the necessary centripetal force.
 
78
0
I have the same problem as Aki, what I did I set initial Energy = final Energy, since energy is conserved, so i end up having : gh = 0.5V^2 , just like what ehild had above. But how I can I get V?
 

Doc Al

Mentor
44,803
1,061
huskydc said:
But how I can I get V?
By applying Newton's 2nd law to the cart as it passes the top of the loop. (As ehild shows.)

Once you get V (or better, once you get [itex]1/2mV^2[/itex]), then apply conservation of energy.
 

ehild

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1,766
There is no friction. The mechanical energy is conserved. If y the cart is at the top of the loop,

(a) [tex]0.5 mv^2=mgh[/tex]

In the loop, the cart performs circular motion. The centripetal force, needed to this motion with speed v results as the sum of both the normal (to the loop) component of gravity and the normal force exerted by the loop. At the top both point vertically downward.

(b) [tex]Fcp=mg+Fn=mv^2/R[/tex]

For safety,

(c) [tex]Fn\ge 0.6 mg[/tex]

Well, I go further.

We express mv^2 from eq. (a) and insert it into eq. (b)

[tex] mv^2= 2mgh [/tex]

[tex]mg+Fn=2mgh/R \rightarrow Fn=mg(2h/R-1)[/tex]

The last condition (c) means the requirement given in the problem, that

...in order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.6 times the weight of the cart.
The weight of the cart is mg. So

[tex]Fn=mg(2h/R-1)\ge 0.6 mg\rightarrow 2h/R-1\ge0.6\rightarrowh/R\ge0.8[/tex]

So the starting height above the top of the loop should be at least 0.8 times the radius of the loop.

Check, please.

ehild
 

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