Q73P - University Physics with Modern Physics

Question

Question: A 42.0-cm-diameter wheel, consisting of a rim and six spokes, is constructed from a thin, rigid plastic material having a linear mass density of \(25.0\;{\rm{g/cm}}\). This wheel is released from rest at the top of a hill 58.0 m high. (a) How fast is it rolling when it reaches the bottom of the hill? (b) How would your answer change if the linear mass density and the diameter of the wheel were each doubled?

Step-by-Step Solution

Answer

(a)26 m/s

Step 1: Given Data

\(\begin{array}{l}{\rm{Diameter}},\;D = 42\;{\rm{cm}} = 0.42 \times {10^{ - 2}}{\rm{m}}\\{\rm{Radius,}}\;r = 0.42 \times {10^{ - 2}}{\rm{m}}\\{\rm{Height,}}\;h = \;58\;{\rm{m}}\end{array}\) 

Step 2: Concept

The relationship between the linear velocity and the angular velocity is \(V = r\omega \).

Where, V is linear velocity and \(\omega \)is angular velocity.

Step 3(a): Determine how fast it is rolling when it reaches the bottom of the hill

Apply conservation of energy,

\(\begin{array}{c}{K_1} + {U_1} = {K_2} + {U_2}\\Mgh + 0 = 0 + \frac{1}{2}M{V^2} + \frac{1}{2}I{\omega ^2} \cdot  \cdot   \cdot  \cdot  \cdot   \cdot \left( 1 \right)\end{array}\)

Where,

\(\begin{array}{c}M = {m_{rim}} + 6{m_{spokes}}\\I = {I_{rim}} + {I_{spokes}} = {m_{rim}}{r^2} + 6\left( {\frac{{{m_{spokes}}{r^2}}}{3}} \right)\end{array}\)

\({m_{rim}} = \lambda \left( {2\pi r} \right),\;{m_{spokes}} = \lambda r\)

Substitute these values in “M” and “I”,

 

\(\begin{array}{c}M = \lambda \left( {2\pi r} \right) + 6\left( {\lambda r} \right)\\ = 2\lambda r\left( {\pi  + 3} \right)\end{array}\)

\(\begin{array}{c}I = \lambda \left( {2\pi r} \right){r^2} + 6\left( {\frac{{\lambda r\left( {{r^2}} \right)}}{3}} \right)\\ = 2\lambda \pi {r^3} + 2\lambda {r^3}\\ = 2\lambda {r^3}\left( {\pi  + 1} \right)\end{array}\)

Substitute M and I value in Equation (1), to find angular velocity \(\omega \), 

\(\begin{array}{c}2r\lambda \left( {\pi  + 3} \right)gh = \frac{1}{2} \times 2r\lambda \left( {\pi  + 3} \right) \times {\left( {r\omega } \right)^2} + \frac{1}{2} \times 2\lambda {r^3}\left( {\pi  + 1} \right) \times {\omega ^2}\\\omega  = \sqrt {\frac{{\left( {\pi  + 3} \right)gh}}{{{r^2}\left( {\pi  + 2} \right)}}} \\ = \sqrt {\frac{{\left( {\pi  + 3} \right) \times 9.8 \times 58}}{{{{\left( {0.21 \times {{10}^{ - 2}}} \right)}^2}\left( {\pi  + 2} \right)}}} \\ = 12400\;{\rm{rad/s}}\end{array}\)

To find the linear velocity,

\(\begin{array}{c}V = r\omega \\ = 0.21 \times {10^{ - 2}} \times 12400\\ = 26\;{\rm{m/s}}\end{array}\)

Hence, the velocity is 26 m/s.