March 27, 2017

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A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls 16 m along the slope?

I got the height by 16sin47=11.7 m
I tried putting the k.e and p.e equations equal......lost!

  • Physics - ,

    Step 1: We know that energy is conserved... so we know that KE(initial)+ PE(initial)= KE(final)+ PE (final).

    Step 2: Since the hoop is initially at rest at the top of the hill, KE (initial) is 0. And after the hoop travels down the hill, all the PE is converted to KE, so PE(final) is equal to 0. So now we have that PE(initial)=KE(final)

    Step 3: Plug in the formulas for PE and KE. PE is equal to mass*gravity*change in height. KE in this problem is KE(rotational)+KE(transitional). KE(rotational)= (1/2)*moment of inertia*angular velocity^2. KE(transitional)= (1/2)*mass*velocity^2.
    So know we have: m*g*change in height= (1/2)*I*w^2+ (1/2)*m*v^2

    Step 4: angular velocity (w) is equal to tangential velocity divided by radius (v/r). So substituting in this value for w, we now have: m*g*change in height= (1/2)*I*(v/r)^2+ (1/2)*m*v^2

    Step 5: The moment of inertia for a hoop is mass*radius^2. So substituting this in for I, we now have: m*g*change in height=(1/2)*m*(r^2)*(v/r)^2+ (1/2)*m*v^2

    Step 6: Simplify the equation! Notice that all the terms are multiplied by a factor of m, so we can pull it out. Also notice that in the KE(rotational) term r^2 cancels out. So now we have: g*change in height= (1/2)*v^2+(1/2)*v^2

    Step 7: Solve for change in height! We know that the slope is 16 meters. And the angle of the slope is 47. sin(theta)=(opposite/hypotenuse)... so the change in height is 16*sin(47). Plugging that in for change in height, we now have: g*16*sin(47)= (1/2)*v^2+(1/2)*v^2

    Step 8: Plug in 9.8 for gravity and solve for v. We get that v= 10.7 m/s!!!!

    Good Luck!!! :)

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