Questions: You stand on a frictional platform that is rotating at 1.3 rev / s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 9.6 kg * m^2. When you pull the weights in toward your body, the moment of inertia decreases to 5.4 kg * m^2. What is the resulting angular speed ωf of the platform? ωf= rev/s What is the change in kinetic energy ΔK of the system? ΔK= J Where did this increase in energy come from? - kinetic energy of the platform - gravity - air resistance - your internal energy - mass of the weights

You stand on a frictional platform that is rotating at 1.3 rev / s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 9.6 kg * m^2. When you pull the weights in toward your body, the moment of inertia decreases to 5.4 kg * m^2.

What is the resulting angular speed ωf of the platform?
ωf=
rev/s

What is the change in kinetic energy ΔK of the system?
ΔK=
J

Where did this increase in energy come from?
- kinetic energy of the platform
- gravity
- air resistance
- your internal energy
- mass of the weights

Transcript text: Question 1 of 3 You stand on a frictional platform that is rotating at $1.3 \mathrm{rev} / \mathrm{s}$. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is $9.6 \mathrm{~kg} \cdot \mathrm{~m}^{2}$. When you pull the weights in toward your body, the moment of inertia decreases to $5.4 \mathrm{~kg} \cdot \mathrm{~m}^{2}$. What is the resulting angular speed $\omega_{\mathrm{f}}$ of the platform? \[ \omega_{\mathrm{f}}= \] $\square$ rev/s What is the change in kinetic energy $\Delta K$ of the system? \[ \Delta K= \] $\square$ J Where did this increase in energy come from? kinetic energy of the platform gravity air resistance your internal energy mass of the weights Question Source: Tipler 6e - Physics For Scientists And Engineers (+Prelectures, VS) Publisher: W.H. Freeman

Solution

Solution Steps

Step 1: Understand the Conservation of Angular Momentum

The problem involves a rotating system where the moment of inertia changes. According to the conservation of angular momentum, the initial angular momentum must equal the final angular momentum if no external torques act on the system. The formula for angular momentum $ L $ is:

\[ L = I \omega \]

where $ I $ is the moment of inertia and $ \omega $ is the angular speed.

Step 2: Apply Conservation of Angular Momentum

Initially, the system has a moment of inertia $ I_i = 9.6 \, \text{kg} \cdot \text{m}^2 $ and an angular speed $ \omega_i = 1.3 \, \text{rev/s} $. When the weights are pulled in, the moment of inertia decreases to $ I_f = 5.4 \, \text{kg} \cdot \text{m}^2 $. Using conservation of angular momentum:

\[ I_i \omega_i = I_f \omega_f \]

Substitute the known values:

\[ 9.6 \times 1.3 = 5.4 \times \omega_f \]

Step 3: Solve for the Final Angular Speed

Solve the equation for $ \omega_f $:

\[ \omega_f = \frac{9.6 \times 1.3}{5.4} \]

Calculate:

\[ \omega_f = \frac{12.48}{5.4} \approx 2.3111 \, \text{rev/s} \]

Step 4: Calculate the Change in Kinetic Energy

The rotational kinetic energy $ K $ is given by:

\[ K = \frac{1}{2} I \omega^2 \]

Calculate the initial and final kinetic energies:

\[ K_i = \frac{1}{2} \times 9.6 \times (1.3)^2 \]

\[ K_f = \frac{1}{2} \times 5.4 \times (2.3111)^2 \]

Calculate each:

\[ K_i = \frac{1}{2} \times 9.6 \times 1.69 = 8.112 \, \text{J} \]

\[ K_f = \frac{1}{2} \times 5.4 \times 5.3421 \approx 14.4197 \, \text{J} \]

The change in kinetic energy $ \Delta K $ is:

\[ \Delta K = K_f - K_i = 14.4197 - 8.112 = 6.3077 \, \text{J} \]

Step 5: Identify the Source of Increased Energy

The increase in kinetic energy comes from the work done by your muscles as you pull the weights inward. This energy is derived from your internal energy.