Physics, 09.03.2020 05:13 AleOfficial101

A.)A merry-go-round has a mass of 317 kg and a radius of 3.36 m. The merry-
go round is spinning such that it completes one revolution every 3.58 s
with a 63.6 kg man standing at its center. (a) What is the moment of
inertia of the merry-go-round if we assume it is a uniform circular disc?
tries: 0
B.)What is the angular velocity of the merry-go-round?
C.)If the person walks straight to the outside of the merry-go-round (i. e.
along a radial line), what is its new angular velocity of the merry-go-
round when he reaches its edge?

Answers: 1

Show answers

Answers

Answer from: littleprinces

A) 1789.4 kg m^2

B) 1.76 rad/s

C) 1.26 rad/s

Explanation:

The moment of inertia of a uniform circular disc rotating about its axis is given by

$I=\frac{1}{2}MR^2$

where

M is the mass of the disc

R is the radius of the disc

For the merry-go-round in this problem, we have:

M = 317 kg is the mass of the disc

R = 3.36 m is its radius

Substituting, we find the moment of inertia:

$I=\frac{1}{2}(317)(3.36)^2=1789.4 kg m^2$

The angular velocity of the merry-go-round is the rate of change of its angular displacement, and it is given by:

$\omega=\frac{2\pi}{T}$

where

$\omega$ is the angular velocity

T is the period of revolution (the time taken for completing 1 revolution)

For the merry-go-round in this problem, it takes 3.58 s to complete one revolution so:

T=3.58 s is the period of revolution

Therefore, the angular velocity is:

$\omega=\frac{2\pi}{3.58}=1.76 rad/s$

When the man walks towards the edge, the angular momentum of the system, must be conserved, since there are no external torques acting on it.

The angurlar momentum is given by:

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Since the angular momentum must remain constant, we can write:

$L_1 =L_2\\I_1\omega_1 = I_2 \omega_2$

where:

I_1=1789.4 kg m^2 is the initial moment of inertia of the system

$\omega_1=1.76 rad/s$ is the initial angular velocity

I_2=I_1+mR^2 is the final moment of inertia, where:

m = 63.6 kg is the mass of the man

R = 3.36 m is the final distance of the man from the axis

Substituting,

I_2=1789.4+(63.6)(3.36)^2=2507.4 kg m^2

And $\omega_2$ is the final angular velocity. Substituting,

$\omega_2=\frac{I_1 \omega_1}{I_2}=\frac{(1789.4)(1.76)}{2507.4}=1.26 rad/s$

Answer from: Quest

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.

Answer from: Quest

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

$v_j = v_1 + v_2 = 3.100 + 1.350 = 4.45 m/s$

now the speed of paul will be same as it is given as he is running on the ground

$v_p = 3.100 m/s$

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

$\delta t = \frac{d}{v_j} - \frac{d}{v_p}$

$\delta t = \frac{2080}{4.450} - \frac{2080}{3.100}$

$\delta t = -203.55 s$

so jerome will reach first and the time difference will be 203.55 s or 3.4 min

Another question on Physics

Physics, 21.06.2019 21:40

Avery hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. the ball rebounds back at essentially the same speed. if the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?

Answers: 3

Answer

Physics, 22.06.2019 11:30

With the simplified model of the eye, what corrective lens (specified by focal length as measured in air) would be needed to enable a person underwater to focus an infinitely distant object? (be careful-the focal length of a lens underwater is not the same as in air! assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. assume that the eyeglasses are 1.90

Answers: 1

Answer

Physics, 22.06.2019 21:00

How is the acceleration of an object in uniform circular motion constant? a - in magnitude only b - in direction only c - in both magnitude and direction d - in magnitude, direction & orientation

Answers: 3

Answer

Physics, 23.06.2019 00:30

Which of the following statements are evidence that gases do not always behave ideally? check all that apply.? a. co2 gas becomes dry ice (solid co2) at 1 atm and –78.5 °c. b. when two gases are mixed, they follow dalton\'s law of partial pressures. c. at 4 k and 1 atm, helium is a liquid. d. it is impossible to compress a gas enough so that it takes up no volume.

Answers: 3

Answer

You know the right answer?

A.)A merry-go-round has a mass of 317 kg and a radius of 3.36 m. The merry-
go round is spinni...

Questions