Mechanics of a ball sliding and rolling

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Homework Statement



A ball of radius r and mass m starts rolling without slipping up a ramp inclined at an angle φ to the horizontal and reaches a maximum height h. Derive an expression for the angular velocity ω that the ball has at the base of the ramp. [Rolling friction may be ignored in this question.]
The ball is initially launched without rotation towards the ramp along a horizontal surface with a constant coefficient of sliding friction μ. The ball slides along the surface, begins to roll and stops slipping before it reaches the ramp. Find an expression for the time ts taken for it to stop sliding in terms of h, g and μ.
By considering the initial velocity v0 and the resistance felt by the ball before it begins to roll, derive an expression relating its initial and final energies in terms of the sliding distance x.


Homework Equations


Kinetic energy of purely rolling body =
[tex]
\frac{\omega^2 I}{2}}
[/tex]


MOI of a solid sphere =
[tex]
\frac{2 M r^2}{5}
[/tex]


The Attempt at a Solution



So I have solved the first part of the question by equating the inital kinetic energy to the final potential energy:
[tex] \frac{\omega^2}{2} = m g h [/tex]

[tex] \omega^2 = \frac{5 g h}{r^2} [/tex]

The second part is where i am having difficulty. Firstly I assume the force due to friction is [itex] \mu m g [/itex]. From here i have tried two routes, setting up the equation of motion and integrating and using conservation of energy, both of which leave me with the problem of needing to know the initial projected velocity to work out the time.

EOM : [itex] m\ddot{x} = - \mu m g [/itex]
[tex] v - v0 = - \mu g t [/tex]

by energy: [itex] \frac{m v_0^2}{2} = \frac{\omega^2 I}{2} + \mu g s [/itex] where s is the distance traveled and is given by the ball before the ramp and can be worked out by the suvat equations.

I am at a lost about how to approach this, plz give me some hint as to which direction i go with it. Thank you for any help. Btw this is my first time posting so i apologize if it isnt set out correctly.
 

Answers and Replies

  • #2
Doc Al
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So I have solved the first part of the question by equating the inital kinetic energy to the final potential energy:
[tex] \frac{\omega^2}{2} = m g h [/tex]

[tex] \omega^2 = \frac{5 g h}{r^2} [/tex]
The ball translates as well as rotates.

The second part is where i am having difficulty. Firstly I assume the force due to friction is [itex] \mu m g [/itex]. From here i have tried two routes, setting up the equation of motion and integrating and using conservation of energy, both of which leave me with the problem of needing to know the initial projected velocity to work out the time.
Consider the angular acceleration.
 

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