Question

A merry-go-round accelerating uniformly from rest achieves itsoperating speed of 2.5rpm in five revolution.

A) What is the magnitude of its angluar acceleration(rad/s^2)?

B) How many seconds does it take the merry-go-round to reachits operating speed?

Answer 1

The student's question requires calculating the angular acceleration of a merry-go-round that accelerates uniformly from rest to an operating speed of 2.5 rpm over five revolutions, and determining the time taken to reach this operating speed.

Step-by-step explanation:

The student has asked two questions about the motion of a merry-go-round:

What is the magnitude of its angular acceleration (rad/s2)?

How many seconds does it take the merry-go-round to reach its operating speed?

To answer part A, we need to find the angular acceleration given the parameters. Since the unit for rotational speed is revolutions per minute (rpm) and we want radians per second, we'll first convert the operating speed to radians per second using the conversion 1 rpm = 2π/60 rad/s. So 2.5 rpm is (2.5 * 2π/60) rad/s. Next, since it's given that the merry-go-round achieves this speed in 5 revolutions, we can use the equation for angular motion under constant acceleration: θ = ω0t + (1/2)αt2, where θ is the angular displacement, ω0 is the initial angular velocity (0 in this case), α is the angular acceleration, and t is the time. Since we're looking for α and we know the merry-go-round makes 5 revolutions, we convert this to radians (5 * 2π) and solve the equation for α, yielding an angular acceleration value.

For part B, using the angular acceleration found in part A, we'll use the relationship ω = ω0 + αt, where ω is the final angular velocity. We solve for time (t) to find how many seconds it takes to reach the operating speed.

Answer 2

(A) The magnitude of its angular acceleration 1.09 x 10⁻³ rad/s²

(B) Time of motion 240.2 s

Step-by-step explanation:

Given;

final angular speed, ωf = 2.5 RPM

angular distance, θ = 5 rev = 5 x 2π = 10π rad

initial angular speed, ωi = 0

final angular speed in rad/s;

`\omega_f = (2.5 \ rev)/(min) * \ (2\pi)/(1 \ rev) * \ (1 \min)/(60 s) = 0.2618 \ rad/s \\`

(A) the magnitude of its angular acceleration(rad/s^2);

`\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\(0.2618)^2 = 0 + (2* 10\pi)\alpha\\\\0.0685 = 20\pi \alpha\\\\\alpha = (0.0685)/(20\pi) \\\\\alpha = 1.09* 10^(-3) \ rad/s^2`

(B) Time of motion;

`\omega_f = \omega_i + \alpha t\\\\0.2618 = 0 + 1.09* 10^(-3) t\\\\t = (0.2618)/(1.09* 10^(-3)) \\\\t = 240.2 \ s`