Ask question

Ask question

All categories

3 years ago

5

A disk-shaped merry-go-round of radius 2.93 m and mass 165 kg rotates freely with an angular speed of 0.691 rev/s . A 62.4 kg pe

rson running tangential to the rim of the merry-go-round at 3.11 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. Calculate the final kinetic energy for this system

2 answers:

Elis [28]3 years ago

6 0

Answer:

The final kinetic energy of the system is 8.58 m/s

Solution:

As per the question:

Radius of merry-go-round, R = 2.93 m

Mass of merry-go-round, m = 165 kg

Angular speed, $\omega = 0.691\ rev/s$

Velocity of the merry-go-around, v = 3.11 m/s

Mass of man, M = 62.4 kg

Now,

To calculate the moment of inertia of the merry-go-round:

$I_{M} = [tex]\frac{1}{2}mR^{2}$

$I_{M} = [tex]\frac{1}{2}\times 165\times (2.93)^{2} = 708.25\ kg.m^{2}$

$I = I_{M} + MR^{2} = 267.85 + 62.4\times (2.93)^{2} = 1243.95\$

Initially, the angular momentum is given by:

$L = MvR + I_{m}\omega$

$L = 62.4\times 3.11\times 2.93 + 708.25\times 0.691\frac{2\pi \ rad/s}{1\ rev/s} = 3643.60\ Js$

The final angular momentum is given by:

$L' = I\omega' = 1243.95\omega'$

where

$\omega'$ = final angular velocity

Now, by conservation of momentum:

Initial momentum = Final momentum

$L' = L$

$1243.95\omega' = 3643.60$

$\omega' = 2.93\ rad/s$

The final linear velocity, v' of the system:

$\omega' = \frac{v'}{R}$

$v' = R\omega' = 2.93\times 2.93 = 8.58\ m/s$

vekshin13 years ago

3 0

Answer:

Explanation:

The problem is related to rotational motion . So we shall find out rotational kinetic energy .

K E = 1/2 x I ω²

ω is the final angular velocity

Moment of inertial of the disk

I ₁ = 1/2 m r²

= .5 x 165 x 2.93²

= 708.25 kgm²

Moment of inertial of the person

I₂ = mr²

= 62.5 x 2.93²

= 536.55 kgm²

ω₂ = v / R

= 3.11 / 2.93 rad /s

At the time of jumping , law of conservation of angular momentum will apply

I₁ ω₁ + I₂ω₂ = (I₁ + I₂)ω

708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω

ω = 0 .85 rad/ s

K E = 1/2 x I ω²

= .5 x ( 708.25 + 536.55 ) ( .85 )²

449.68 J

You might be interested in

two mass Ma=2kg and Mb=5kg on incline are connected together by string as shown below the coefficient of kinetic friction betwee

aniked [119]

Answer:

yourjejejrjrjrjr mother

5 0

3 years ago

The primary source of evidence proposed by many scientist to support the theory of an ancient earth is ____ dating.

vova2212 [387]

It depends on your definition of “ancient.” Radiometric dating using Carbon-14 can reliably date back to about 50,000 years, uranium-lead or lead-lead dating can date back multiple millions, potassium-argon dating can reach 1.5 billion, and rubidium-strontium can reach 50 billion (nearly 4x the age of the universe). It depends on the context in which this question is being asked.

7 0

3 years ago

When you see yourself in a plane mirror, the image is always:

Charra [1.4K]

It’s c “ the same size as you are”

6 0

3 years ago

Which two possible explanations does the video give for why white light changes into colored light? Color is an impurity added t

Alex777 [14]

Answer:

White light contains seven colors, which are separated by the prism.

White light reacts with chemicals in the air to produce seven colors.

Explanation:

It was first observed by Sir Issac Newton, that when white light is passed through a prism, an elongated, coloured patch of light is obtained on a screen placed behind the prism. The seven colours obtained constitute the spectrum of white light.

In nature, white light is separated into its constituent wavelengths when white light interacts with substances in the atmosphere.

3 0

4 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

4 years ago