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A homogeneous sphere starts from rest at the upper end of the track pictured, th

ID: 1501261 • Letter: A

Question

A homogeneous sphere starts from rest at the upper end of the track pictured, then rolls without slipping until it rolls off the right hand end (which is horizontal), a) Draw two freebody diagrams, one during the time it is rolling and one after it has rolled off the right hand end. Write down your second law equations, putting in all your forces. You should have three equations for the first and one for the second (why?) b) Is the energy of the ball conserved during the rolling, after it has rolled off the end, and why or why not? c) Is the linear momentum of the ball conserved during either or both of these times, and why or why not? d) Is the angular momentum conserved during these times, and why or why not? e) Find where the ball hits the

Explanation / Answer

A)A homogeneous sphere starts from rest .It rolls without slipping.

free body diagram at the top(at height H)

gravitational force is acting downward, after it reached the right end also gavitational force acting down ward

but at the top only potential energy is acting on the object. but at the bottom it g=has kinetic energy and rotational kinetic energy

At the top . it has only potential energy U1 = mgH

At the bottom it has kinetic energy +rotational kinetic energy + potential energy = 1/2*m*v^2 + 1/2*I*^2 + m*g*h

b) There are no other forces acting on the object , so that energy will be conserved.Because there is no loss of energy. No external forces are acting on the object , so that linear momentum and angular momentum are conserved.

e) From conservation of energy we have U1 = K2 + U2

here K2 includes both translation and rotational kinetic energy

So m*g*H = 1/2*m*v^2 + 1/2*I*^2 + m*g*h

Now I = 2/5*m*r^2 and = v/r

so m*g*(H - h) = 1/2*m*v^2 + 1/2*2/5*m*r^2*v^2/r^2 = 1/2*m*v^2*(1 + 2/5)

so v = sqrt(2*g*(H - h)/1.4)

time to reach this distance t = sqrt(2*h/g)

the horizontal distance = v*t

=  sqrt(2*g*(H - h)/1.4) * sqrt(2*h/g)

= sqrt(4*h(H-h)/1.4)

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