Conservation of energy

Classical Mechanics Level 2

A ball of mass m = 5 m=5 kg is rolled, from rest, down a metal track, starting from point A as shown. Point A is 10 10 meters above the ground, and the radius of the loop in the track is R = 1 R= 1 meters. How fast is it moving at the point D?


The answer is 14.

Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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2 solutions

It has kinetic energy 1/2 m vD2, but zero potential energy since it is at ground level (h=0 meters). So its total energy at D, ED is

E D = 1 / 2 m v D 2 ED=1/2 mvD2

Equation 8

But since energy is conserved, Ei=ED, or

m g h A = 1 / 2 m v D 2 mghA=1/2 mvD2

Note that the m's cancel, and we can solve for v to get

v D = 2 g h A vD= 2ghA or the speed at point D is vD=14 m/s

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Note, to display Latex inline, you can use normal brackets i.e. \ ( \ ) (with spaces removed). Using square brackets will make your equations take up a new line, which could be disruptive to the flow. I've edited your problem, and it is looks better. You can edit it to see what I did.

Calvin Lin Staff - 7 years, 2 months ago

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Thanks

Mardokay Mosazghi - 7 years, 2 months ago

Surely the re-arrangement is v = 2 g h v = \sqrt{2gh} . Anyway, the problem specifies that the ball is rolling, so assuming that it is a uniform sphere and there is no slipping, the kinetic energy at D actually equals 7 10 m v 2 \frac{7}{10}m{v}^2 .

A K - 7 years, 1 month ago

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yes, I also think so. If the ball is rolling, it has rotation kinetic energy. So the equation will be m g h = 1 2 m v 2 + 1 2 I ω 2 mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega ^2

then, v = 10 g h 7 v = \sqrt\frac {10gh}{7}

Harianto Wibowo - 7 years, 1 month ago

i am agree to every one who solved it by considering rolling because it is mentioned in problem.......plz change the answer to improve th quality of problem..

Amit Sharma - 7 years, 1 month ago

I took g=10ms^-2 and calculated the answer 20 but it said it was wrong, so I checked this solution. Please do mention the value of g otherwise confusions can happen

Saswata Dasgupta - 6 years, 10 months ago

What will be the velocity at E and F?

Kaushik Chandra - 3 years, 11 months ago
Harpinder Singh
Jun 19, 2014

At point A , potential energy=mgh=5×9.8×10=490N At point D, Potential Energy=Kinetic Energy, 490=1/ 2×M×V^2 V^2=980/5=196 V=14 m/s

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