A solid homogeneous sphere of mass M = 4.00 kg is released from rest at the top of an incline of height H=1.35 m and rolls without slipping to the bottom. The ramp is at an angle of θ = 28.7o to the horizontal.
figure: http://i.imgur.com/lI9Cy.gif
(a)Calculate the speed of the sphere's CM at the bottom of the incline.
(b)Determine the rotational kinetic energy of the sphere at the bottom of the incline.
figure: http://i.imgur.com/lI9Cy.gif
(a)Calculate the speed of the sphere's CM at the bottom of the incline.
(b)Determine the rotational kinetic energy of the sphere at the bottom of the incline.
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As the sphere moves down the ramp, it has two types of kinetic energy, linear and rotational.
The total kinetic energy of the sphere = Liner KE + Rotational KE
Linear KE = ½ * mass * (linear velocity)^2
Rotational KE = ½ * moment of inertia * (angular velocity)^2
moment of inertia of solid sphere = 2/5 * mass * radius^2
angular velocity = linear velocity ÷ radius
Rotational KE = ½ * (2/5 * m * r^2) * (v/r)^2 = 1/5 * m * v^2
The total kinetic energy of the sphere = ½ * m * v^2 + 1/5 * m * v^2 = 0.7 * m * v^2
The component of the weight, which is parallel to the slope of the ramp, causes the linear acceleration of the sphere as it moves down the ramp. The friction force causes the sphere to rotate as it moves down the ramp. So, the friction force causes the angular acceleration of the sphere as it moves down the ramp.
This means the sum of the work done by the force parallel and the work done by the friction force is equal to the initial potential energy of the sphere.
The work done by the force parallel = Linear KE
The work done by the friction force = Rotational KE
Total work = Linear KE + Rotational KE = 0.7 * m * v^2
Initial PE = m * g * h
0.7 * m * v^2 = m * g * h
Divide both sides by (0.7 * m)
v^2 = g * h ÷ 0.7
Linear velocity of sphere at bottom of ramp = √(g * h ÷ 0.7)
g = 9.8 m/s^2
h = 1.35 m
Linear velocity of sphere at bottom of ramp = √(9.8 * 1.35 ÷ 0.7) ≈ 4.347m/s
Final linear KE = ½ * m * v^2 = ½ * 4 * (9.8 * 1.35 ÷ 0.7) = 37.8 J
Final rotational KE = 1/5 * m * v^2 = 1/5 * 4 * (9.8 * 1.35 ÷ 0.7) = 15.12 J
Total Final KE = 37.8 + 15.12 = 52.92 J
Initial PE = m * g * h = 4 * 9.8 * 1.35 = 52.92 J
Total Final KE = Initial PE
When a ball rolls, without slipping, the final kinetic energy of the ball is equal to the initial potential energy of the ball.
This means energy is conserved!
Energy is conserved, because the friction force actually helped the ball move down the ramp. The friction force caused the ball to rotate as it moved down the ramp.
Rolling objects are more efficient than sliding objects. That is why we put wheels on objects, which we want to move!!
I hope this helps you understand how linear and rotational motion interact as an object rolls down an inclined surface.
The total kinetic energy of the sphere = Liner KE + Rotational KE
Linear KE = ½ * mass * (linear velocity)^2
Rotational KE = ½ * moment of inertia * (angular velocity)^2
moment of inertia of solid sphere = 2/5 * mass * radius^2
angular velocity = linear velocity ÷ radius
Rotational KE = ½ * (2/5 * m * r^2) * (v/r)^2 = 1/5 * m * v^2
The total kinetic energy of the sphere = ½ * m * v^2 + 1/5 * m * v^2 = 0.7 * m * v^2
The component of the weight, which is parallel to the slope of the ramp, causes the linear acceleration of the sphere as it moves down the ramp. The friction force causes the sphere to rotate as it moves down the ramp. So, the friction force causes the angular acceleration of the sphere as it moves down the ramp.
This means the sum of the work done by the force parallel and the work done by the friction force is equal to the initial potential energy of the sphere.
The work done by the force parallel = Linear KE
The work done by the friction force = Rotational KE
Total work = Linear KE + Rotational KE = 0.7 * m * v^2
Initial PE = m * g * h
0.7 * m * v^2 = m * g * h
Divide both sides by (0.7 * m)
v^2 = g * h ÷ 0.7
Linear velocity of sphere at bottom of ramp = √(g * h ÷ 0.7)
g = 9.8 m/s^2
h = 1.35 m
Linear velocity of sphere at bottom of ramp = √(9.8 * 1.35 ÷ 0.7) ≈ 4.347m/s
Final linear KE = ½ * m * v^2 = ½ * 4 * (9.8 * 1.35 ÷ 0.7) = 37.8 J
Final rotational KE = 1/5 * m * v^2 = 1/5 * 4 * (9.8 * 1.35 ÷ 0.7) = 15.12 J
Total Final KE = 37.8 + 15.12 = 52.92 J
Initial PE = m * g * h = 4 * 9.8 * 1.35 = 52.92 J
Total Final KE = Initial PE
When a ball rolls, without slipping, the final kinetic energy of the ball is equal to the initial potential energy of the ball.
This means energy is conserved!
Energy is conserved, because the friction force actually helped the ball move down the ramp. The friction force caused the ball to rotate as it moved down the ramp.
Rolling objects are more efficient than sliding objects. That is why we put wheels on objects, which we want to move!!
I hope this helps you understand how linear and rotational motion interact as an object rolls down an inclined surface.