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A boy sitting on a merry-round make 200 ...

A boy sitting on a merry-round make 200 revolutions per minute . Calculate the angular speed of the merry-go-round.

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To solve the problem of calculating the angular speed of a merry-go-round given that a boy makes 200 revolutions per minute, we can follow these steps: ### Step 1: Understand the relationship between revolutions and radians 1 revolution is equal to \(2\pi\) radians. Therefore, if the boy makes 200 revolutions, the total angle in radians can be calculated as: \[ \text{Total angle in radians} = \text{Number of revolutions} \times 2\pi = 200 \times 2\pi \] ### Step 2: Convert revolutions per minute to radians per minute Now, we can calculate the total angle in radians: \[ \text{Total angle in radians} = 200 \times 2\pi = 400\pi \text{ radians} \] ### Step 3: Convert time from minutes to seconds Since the angular speed is typically expressed in radians per second, we need to convert the time from minutes to seconds. There are 60 seconds in a minute, so: \[ \text{Time in seconds} = 1 \text{ minute} = 60 \text{ seconds} \] ### Step 4: Calculate angular speed Angular speed (\(\omega\)) is defined as the total angle traveled divided by the time taken. Therefore: \[ \omega = \frac{\text{Total angle in radians}}{\text{Time in seconds}} = \frac{400\pi \text{ radians}}{60 \text{ seconds}} \] ### Step 5: Simplify the expression Now we simplify the expression: \[ \omega = \frac{400\pi}{60} = \frac{20\pi}{3} \text{ radians per second} \] ### Final Answer The angular speed of the merry-go-round is: \[ \omega = \frac{20\pi}{3} \text{ radians per second} \] ---

To solve the problem of calculating the angular speed of a merry-go-round given that a boy makes 200 revolutions per minute, we can follow these steps: ### Step 1: Understand the relationship between revolutions and radians 1 revolution is equal to \(2\pi\) radians. Therefore, if the boy makes 200 revolutions, the total angle in radians can be calculated as: \[ \text{Total angle in radians} = \text{Number of revolutions} \times 2\pi = 200 \times 2\pi \] ...
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Knowledge Check

  • A playground merry-go-round of radius R = 2.00 has a moment of inertia I = 250 kg.m^2 and is rotating 10.0 rev//min about a frictionless, vertical axle. Facing axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. The new angular speed of the merry-go-round is.

    A
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    B
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    C
    `7.14 rev//min`
    D
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  • The diameter of a flywheel is 1.2 m and it makes 900 revolutions per minute. Calculate the acceleration at a point on its rim

    A
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    `270m//s^2`
    C
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    D
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  • A child is standing on the edge of a merry-go-round that has the shape of a disk, as shown in the figure. The mass of the child is 40 kg. The merry-go-round has a mass of 200kg and a radius of 2.5 m, and it is rotating with an angular velocity of omega=2.0 radians per second. The child then walks slowly towards the centre of the merry-go-round. What will be the final angular velocity of the merry-go-round when the child reaches the centre? (The size of the child can be neglected)

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    C
    2.4 rad/s
    D
    2.8 rad/s
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