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10 July, 16:24

A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why

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  1. 10 July, 16:47
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    Rolling case achieves greater height than sliding case

    Step-by-step explanation:

    For sliding ball:

    - When balls slides up the ramp the kinetic energy is converted to gravitational potential energy.

    - We have frictionless ramp, hence no loss due to friction. So the entire kinetic energy is converted into potential energy.

    - The ball slides it only has translational kinetic energy as follows:

    ΔK. E = ΔP. E

    0.5*m*v^2 = m*g*h

    h = 0.5v^2 / g

    For rolling ball:

    - Its the same as the previous case but only difference is that there are two forms of kinetic energy translational and rotational. Thus the energy balance is:

    ΔK. E = ΔP. E

    0.5*m*v^2 + 0.5*I*w^2 = m*g*h

    - Where I: moment of inertia of spherical ball = 2/5 * m*r^2

    w: Angular speed = v / r

    0.5*m*v^2 + 0.2*m*v^2 = m*g*h

    0.7v^2 = g*h

    h = 0.7v^2 / g

    - From both results we see that 0.7v^2/g for rolling case is greater than 0.5v^2/g sliding case.
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