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IrinaK [193]
3 years ago
13

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

heground and slides along a frictionless track. The car encounters aloop of radius R, shown. Assume that the initial height h is greatenough so that the car never losses contact with the track.
Find an expression for the kinetic energy of the car at the top ofthe loop (express in terms of m,g,h,R).

Find the minimum initial height h at which the car can be releasedthat still allows the car to stay in contact with the track at thetop of the loop (express in terms of R).
Physics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

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Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

  • When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
  • This force is not a new force, is just the net force aiming to the center of the circle.
  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
  • Replacing (2) in (1), and solving for Fn, we get:

       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g}  (4)

  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

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2 years ago
Please help ASAP <3 thanks
Vlad [161]
1.false
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4.true
5.false/not entirely sure
6.true
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bixtya [17]
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The equation is
E = h f
Simply substitute and solve for f which is the frequency
f = </span>2.64 × 10-21 J / 6.6 × 10-34 J s
f = <span>4.00 × 1012<span> hertz</span></span>
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