Physics, 14.01.2021 16:20 deelashasharma

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 passes its lowest point. (a) What is the speed when thestring is at 60# to the vertical? (b) What is the greatest angle withthe vertical that the string will

Answers: 1

Show answers

Answers

Answer from: highlander4215

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:

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

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

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

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:

$\theta_{max}=79.34^{o}$

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a

Answer from: Quest

v=ir = 3x15 =45 v

Answer from: Quest

i think it's 35 meters

Another question on Physics

Physics, 21.06.2019 17:00

Heat from the sun reaches earth to a. radiation b. convection c. conduction

Answers: 2

Answer

Physics, 22.06.2019 06:30

=force × distance a. work b. velocity c. pressure d. momentum

Answers: 1

Answer

Physics, 22.06.2019 18:00

Cells in the nervous system have a potential difference of 70 mv across the cell membrane separating the interior of the cell from the extracellular fluid. this potential difference is maintained by ion pumps that move charged ions across the membrane. is this an emf? select the correct answer and explanation. 1)no. the ion pumps cannot separate charges; thus, they cannot create a potential difference. 2)yes. the ion pumps cannot separate charges, but they still can create a potential difference. 3)yes. the ion pumps can actively separate charge; thus, they can create a potential difference. 4)no. the ion pumps can separate charges, but they cannot create a potential difference.

Answers: 1

Answer

Physics, 23.06.2019 01:30

Which statements did both aristotle and ptolemy assume? check all that apply.

Answers: 1

Answer

You know the right answer?

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a...

Questions