A student of mass \(52 \mathrm{~kg}\) wants to measure the mass of a playground merry-go-round, which consists of a solid metal disk of radius \(R=1.5 \mathrm{~m}\) that is mounted in a horizontal position on a low-friction axle. She tries an experiment: She runs with speed \(v=6.8 \mathrm{~m} / \mathrm{s}\) toward the outer rim of the merry-go-round and jumps on to the outer rim, as shown in the figure. The merry-go-round is initially at rest before the student jumps on and rotates at \(1.3 \mathrm{rad} / \mathrm{s}\) immediately after she jumps on. You may assume that the student's mass is concentrated at a point. a) What is the mass of the merry-go-round? b) If it takes 35 s for the merry-go-round to come to a stop after the student has jumped on, what is the average torque due to friction in the axle? c) How many times does the merry-go-round rotate before it stops, assuming that the torque due to friction is constant?

Step by step solution

01

a) Finding the mass of the merry-go-round

Using the conservation of angular momentum, we can write the equation as: Initial angular momentum = Final angular momentum m_s * v * R = (m_s + M) * I * ω where m_s is the mass of the student, v is the student's speed, R is the merry-go-round radius, I is the moment of inertia of the merry-go-round, M is the mass of the merry-go-round, and ω is the angular speed. It is given that student's mass is concentrated at a point so we can ignore her moment of inertia. The moment of inertia of a solid disk is: \( I = \frac{1}{2} M R^2 \) Solving the equation for the mass of the merry-go-round (M), we have: \( M = \frac{2 m_s v R - m_s R^2 \omega}{R^2 \omega} \) Now, substitute the values: \( M = \frac{2 * 52 * 6.8 * 1.5 - 52 * 1.5^2 * 1.3}{1.5^2 * 1.3} \) Calculating the above expression, we get: M ≈ 254.62 kg So, the mass of the merry-go-round is approximately 254.62 kg.

02

b) Finding the average torque due to friction in the axle

The change in angular momentum of the merry-go-round will be equal to the torque multiplied by the time it takes to come to a stop, which can be written as: ΔL = τ * t The initial angular momentum is equal to (m_s + M) * I * ω, and the final angular momentum is 0 because it comes to a stop. Therefore, ΔL = (m_s + M) * I * ω Calculating the torque τ, we have: τ = ΔL / t Substitute the values and compute the torque: τ = [(52 + 254.62) * (1/2) * 254.62 * 1.5^2 * 1.3] / 35 Calculating the above expression, we get: τ ≈ 30.40 Nm So, the average torque due to friction in the axle is approximately 30.40 Nm.

03

c) Finding the number of rotations before the merry-go-round stops

The work done by frictional torque to stop the merry-go-round equals the initial rotational kinetic energy. So, we can write the equation as: τ * θ = (1/2) * (m_s + M) * I * ω^2 Solving the equation for θ (the number of radians turned before stopping), we have: θ = \(\frac{(m_s + M) * I * \omega^2}{2 * \tau}\) Now, substitute the values: θ = \(\frac{(52 + 254.62) * (1/2) * 254.62 * 1.5^2 * 1.3^2}{2 * 30.40} \) Calculating the above expression, we get: θ ≈ 184.67 rad To find the number of rotations (n), divide the number of radians by 2π: n = θ / (2 * π) Calculating the above expression, we get: n ≈ 29.4 So, the merry-go-round rotates approximately 29.4 times before coming to a stop.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!