After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of = 3.18 m/s.
To reach the rack, the ball rolls up a ramp that rises through a vertical distance of = 0.510 m. What is the linear speed of the ball when it reaches the top of the ramp?
(in m/s)
To find the linear speed of the ball when it reaches the top of the ramp, we can use a principle called conservation of mechanical energy. The total mechanical energy of the ball is conserved as it rolls up the ramp.
The mechanical energy of the ball is the sum of its kinetic energy (KE) and potential energy (PE). At the bottom of the ramp, all of the energy is in the form of kinetic energy, and at the top of the ramp, all of the energy is in the form of potential energy.
The kinetic energy (KE) of an object is given by the equation KE = (1/2) * mass * velocity^2.
The potential energy (PE) of an object is given by the equation PE = mass * gravity * height.
In this case, the mass of the ball and the gravitational acceleration (gravity) are not given, but we can assume they are constant.
Since the ball rolls without slipping, the linear speed of the ball (velocity) can be calculated using the equation v = omega * r, where omega is the angular speed and r is the radius of the ball.
To solve this problem, we need to find the relationship between the linear speed, angular speed, and radius of the ball.
Let's denote the angular speed as omega and the radius of the ball as R. The linear speed of the ball is related to the angular speed by the formula v = omega * R.
Given that the linear speed of the ball is 3.18 m/s, we can substitute this value into the equation v = omega * R to solve for omega.
Solving for omega, we have omega = v / R.
Now, let's find the potential energy (PE) of the ball at the top of the ramp.
PE = mass * gravity * height
Substituting the given values, we have PE = m * g * h, where m is the mass of the ball, g is the gravitational acceleration, and h is the vertical distance the ramp rises.
The total mechanical energy (TE) of the ball is given by the sum of the kinetic energy and potential energy:
TE = KE + PE
Since the total mechanical energy is conserved, TE at the bottom of the ramp is equal to TE at the top:
KE_bottom + PE_bottom = KE_top + PE_top
At the bottom of the ramp, the ball has only kinetic energy, so KE_bottom = (1/2) * m * v^2.
At the top of the ramp, the ball has only potential energy, so PE_top = m * g * h.
Substituting these values into the conservation of energy equation, we have:
(1/2) * m * v_bottom^2 + m * g * h_bottom = 0 + m * g * h_top
Now, we can solve for the final linear speed (v_top) by rearranging the equation:
v_top = sqrt(2 * g * (h_top - h_bottom) + v_bottom^2)
Substituting the given values, we have:
v_top = sqrt(2 * 9.8 m/s^2 * (0.510 m - 0) + (3.18 m/s)^2)
Evaluating the expression, the linear speed of the ball when it reaches the top of the ramp is approximately 3.37 m/s.