A solid, uniform disk of radius 0.250\ m and mass 60.4\ kg rolls down a ramp of length 4.60\ m...

Question:

A solid, uniform disk of radius {eq}0.250\ m {/eq} and mass {eq}60.4\ kg {/eq} rolls down a ramp of length {eq}4.60\ m {/eq} that makes an angle of {eq}18^{\circ} {/eq} with the horizontal. The disk starts from rest from the top of the ramp. (a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp. m/s (b) Find the angular speed of the disk at the bottom of the ramp. rad/s

Conservation of energy:

The transformation of energy from a system of bodies to another system of bodies without destroying is known as the conservation of energy. Most of the energy of the moving converted to heat energy while it stops.

Answer and Explanation:


Given data

  • The value of the radius of the disc is {eq}{r_{disc}} = 0.250\;{\rm{m}} {/eq}
  • The value of the mass of the disc is {eq}{m_{disc}} = 60.4\;{\rm{kg}} {/eq}
  • The value of the length of the ramp is {eq}l = 4.60\;{\rm{m}} {/eq}
  • The value of the angle of inclination is {eq}\phi = 18^\circ {/eq}


The expression for the force due to gravity is,

{eq}{F_g} = {m_{disc}}gl\sin \phi {/eq}


The expression for the force due to angular velocity is,

{eq}\begin{align*} {F_a} &= \dfrac{1}{2}{I_c}{\omega ^2}\\ &= \dfrac{1}{2}\left( {\dfrac{1}{2}{m_{disc}}{r_{disc}}^2} \right){\left( {\dfrac{{{v_{disc}}}}{{{r_{disc}}}}} \right)^2} \end{align*} {/eq}


The expression for the force due to velocity is,

{eq}{F_v} = \dfrac{1}{2}\left( {{m_{disc}}{v_{disc}}^2} \right) {/eq}


From the law of conservation of energy,

{eq}\begin{align*} {F_g} &= {F_a} - {F_v}\\ {m_{disc}}gl\sin \phi &= \left( {\dfrac{1}{2}\left( {\dfrac{1}{2}{m_{disc}}{r_{disc}}^2} \right){{\left( {\dfrac{{{v_{disc}}}}{{{r_{disc}}}}} \right)}^2}} \right) - \left( {\dfrac{1}{2}\left( {{m_{disc}}{v_{disc}}^2} \right)} \right) \end{align*} {/eq}


Substitute the value in the above equation.

{eq}\begin{align*} \left( {60.4} \right)\left( {9.81} \right)\left( {4.60} \right)\sin 18^\circ &= \left( {\dfrac{1}{2}\left( {\dfrac{1}{2}\left( {60.4} \right){{\left( {0.250} \right)}^2}} \right){{\left( {\dfrac{{{v_{disc}}}}{{0.250}}} \right)}^2}} \right) - \left( {\dfrac{1}{2}\left( {\left( {0.640} \right){{\left( {0.250} \right)}^2}} \right)} \right)\\ {v_{disc}} &= 7.4686\;{\rm{m/s}} \end{align*} {/eq}


Thus, the value of the velocity of the disk?s centre of mass is{eq}7.4686\;{\rm{m/s}} {/eq}


(b)

The expression for the angular velocity is,

{eq}{\omega _{ang}} = \dfrac{{{v_{disc}}}}{{{r_{disc}}}} {/eq}


Substitute the value in the above equation.

{eq}\begin{align*} {\omega _{ang}} &= \dfrac{{7.4686}}{{0.250}}\\ &= 29.8744\;{\rm{rad/s}} \end{align*} {/eq}


Thus, the value of the angular velocity of the disk is {eq}29.8744\;{\rm{rad/s}} {/eq}


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