1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Snowcat [4.5K]
3 years ago
12

A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an ang

ular momentum of 10,000 (kg m 2 radians/s) and its outer edge has a speed of 2 m/s. What is its radius, in m
Physics
1 answer:
lapo4ka [179]3 years ago
5 0

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

<u>r = 20 m</u>

You might be interested in
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
What is the SI (metric) unit of FORCE?<br><br> A. meter<br> B. newton
Tresset [83]

What is the SI (metric) unit of FORCE?

  • B. newton

with symbol ( N )

All the best !

4 0
2 years ago
Read 2 more answers
What problems do pathogens cause in our society?
kotegsom [21]

Pathogens are microorganisms that cause infectious diseases. Bacteria and viruses are the main pathogens in the world today.

The major problem that pathogen cause in society is that they are the agents of infection and disease. 

An infection results when a pathogen invades the body and begins growing within a host.  A disease results when tissue function is impaired due to an infection.

Due to diseases, affected people can not work  which negatively affects the economy, and the sick may end up dying causing a lot of grief, and robbing society of a father, a mother or child.

The government has to put money aside for health care which would otherwise have been used for other things to help the society.

3 0
3 years ago
A car travel with a constant velocity of 20.5m/s for 20seconds what distance does it cover in this time ?​
prohojiy [21]

Answer:

410 m

Explanation:

Given:

v₀ = 20.5 m/s

a = 0 m/s²

t = 20 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20.5 m/s) (20 s) + ½ (0 m/s²) (20 s)²

Δx = 410 m

8 0
3 years ago
4)
Citrus2011 [14]

Answer:

net force is positive downward..B

6 0
3 years ago
Other questions:
  • Which of earths layers is made up of liquid metal
    10·2 answers
  • In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w
    13·1 answer
  • A 300-kg truck moving rightward with a speed of 5 km/hr collides head on with a 1000kg car moving leftward with a speed of 100 k
    7·1 answer
  • What form of energy are microwaves?
    7·1 answer
  • Help with all of them plsss
    6·1 answer
  • Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele
    8·1 answer
  • How do objects at rest and in motion respond in the presence of an external, unbalanced force?
    13·1 answer
  • heres free wallpapers and pnts! :)) (phone and PC and chromebook wallpapers) (credits to those who made them !)
    5·1 answer
  • The equation most often associated with Newton's Second Law of Motion is<br> Answer here
    8·2 answers
  • Only the smartest person in science can help me right now...
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!