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mixer [17]
4 years ago
8

HELP!! Worth a lot of points :) (Physics question)

Physics
2 answers:
Temka [501]4 years ago
8 0
My answer would be B.
Source:
i had this question before.
Hope that this helps you! =)
nadezda [96]4 years ago
6 0
It's B if that helps at all :) (I'm like, 98% sure that it's totally B!
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A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass
arlik [135]

Answer:

a) v_{60^{o}} =4.98 m/s

b) \theta_{max}=79.34^{o}

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

K_{lowest}=K_{60^{o}}+U_{60^{0}}

So now, we can substitute the respective equations for kinetic and potential energy so we get:

\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}

we can divide both sides of the equation into the mass of the pendulum so we get:

\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}

and we can multiply both sides of the equation by 2 to get:

v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}

so we can solve this for v_{60^{o}}. So we get:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

cos \theta = \frac{y}{4m}

so we solve for y to get:

y = 4cos \theta

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

h_{60^{o}}=4m-2m

h_{60^{o}}=2m

So now we can substitute the data in the velocity equation we got before:

v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}

v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}

so

v_{60^{o}}=4.98 m/s

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

K_{lowest}+U_{lowest}=K_{max}+U_{max}

In this case, we know that U_{lowest} will be zero and K_{max} will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

K_{lowest} =U_{max}

And now we substitute for the respective kinetic energy and potential energy equations.

\frac{1}{2}mv_{lowest}^{2}=mgh_{max}

again, we can divide both sides of the equation into the mass, so we get:

\frac{1}{2}v_{lowest}^{2}=gh_{max}

and solve for the height:

h_{max}=\frac{v_{lowest}^{2}}{2g}

and substitute:

h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}

to get:

h_{max}=3.26m

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

cos \theta_{max}= \frac{0.74}{4}

\theta_{max}=cos^{-1}(\frac{0.74}{4})

so we get:

\theta_{max}=79.34^{o}

5 0
3 years ago
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