A spherical bowling ball with mass m = 3.7 kg and radius R = 0.103 m is thrown down the lane with an initial speed of v = 8.9 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.31. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.
a. What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?
b. What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?
c. How long does it take the bowling ball to begin rolling without slipping?
d. How far does the bowling ball slide before it begins to roll without slipping?
e. What is the magnitude of the final velocity?
f. After the bowling ball begins to roll without slipping, compare the rotational and translational kinetic energy of the bowling ball:
(i) KErot < KEtran
(ii) KErot = KEtran
(iii) KErot > KEtran
To solve this problem, we can use the equations of motion and the conditions for rolling without slipping. Let's go step by step to find the answers to each question.
a. To find the magnitude of the angular acceleration, we can use the equation:
α = a / R,
where α is the angular acceleration and a is the linear acceleration. Since the ball is initially sliding, the linear acceleration can be found using Newton's second law:
F_friction = μ * F_normal,
F_friction = μ * m * g,
a = F_friction / m = μ * g,
where F_friction is the frictional force, F_normal is the normal force, m is the mass of the ball, μ is the coefficient of kinetic friction, and g is the acceleration due to gravity.
Substituting the values, we get:
a = μ * g = 0.31 * 9.8 m/s²,
Now, substituting the value of a into the equation for angular acceleration:
α = a / R = (0.31 * 9.8 m/s²) / 0.103 m.
Calculating this will give you the magnitude of the angular acceleration.
b. The magnitude of the linear acceleration is already calculated in part a:
a = 0.31 * 9.8 m/s².
c. To find the time it takes for the ball to begin rolling without slipping, we can use the equation:
t = v / a,
where t is the time, v is the initial speed, and a is the linear acceleration. Substituting the values, we get:
t = 8.9 m/s / (0.31 * 9.8 m/s²).
Calculating this will give you the time taken for rolling without slipping.
d. To find the distance the ball slides before it begins to roll without slipping, we can use the equation:
d = v * t + (1/2) * a * t²,
where d is the distance, v is the initial speed, t is the time, and a is the linear acceleration. Substituting the values, we get:
d = 8.9 m/s * t + (1/2) * (0.31 * 9.8 m/s²) * t².
Calculating this will give you the distance the ball slides.
e. After the ball begins to roll without slipping, the final velocity can be calculated using the equation:
v_final = ω * R,
where v_final is the final velocity, ω is the angular velocity, and R is the radius of the ball. Since the ball is rolling without slipping, the angular velocity can be calculated using the equation:
ω = α * t,
where α is the angular acceleration and t is the time taken for rolling without slipping. Substituting the values, we get:
ω = α * t = (0.31 * 9.8 m/s²) * t.
Finally, substituting the values of ω and R into the equation for final velocity:
v_final = (0.31 * 9.8 m/s²) * t * 0.103 m.
Calculating this will give you the magnitude of the final velocity.
f. To compare the rotational and translational kinetic energy, we can use the equations:
KE_rot = (1/2) * I * ω²,
KE_tran = (1/2) * m * v_final²,
where KE_rot is the rotational kinetic energy, KE_tran is the translational kinetic energy, I is the moment of inertia of the ball, ω is the angular velocity, m is the mass of the ball, and v_final is the final velocity.
Comparing these equations, we can make the following conclusion:
Since both kinetic energies are proportional to the square of their respective velocities, and the ball is rolling without slipping, the moment of inertia is related to the mass and the radius (I = (2/5) * m * R²). Therefore, in this case:
KE_rot = KE_tran. (ii)
This means that the rotational and translational kinetic energies are equal.
To solve these questions, we'll use the concepts of rotational motion, linear motion, and the relationship between them. Let's go through each question step by step.
a. The angular acceleration of the bowling ball can be found using the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. In this case, the torque is caused by the frictional force acting on the ball. The frictional force is given by f = μN, where μ is the coefficient of kinetic friction and N is the normal force. The normal force, in this case, is equal to the weight of the ball, which is given by N = mg. The moment of inertia for a solid sphere rotating about its center is given by I = (2/5)mr^2, where m is the mass of the ball and r is the radius. So, we have τ = (2/5)mR^2α. Since the linear acceleration is zero initially, the angular acceleration is the same as the linear acceleration. Therefore, the magnitude of the angular acceleration is given by α = τ / I = (5/2)(μmgR) / (2/5)mR^2 = (5/2)μg.
b. The magnitude of the linear acceleration of the bowling ball can be found using the equation f = ma. In this case, the net force on the ball is due to the frictional force and is given by f = μN = μmg. So, we have μmg = ma, where a is the linear acceleration. Therefore, the magnitude of the linear acceleration is given by a = μg.
c. The ball will begin to roll without slipping when the static friction force between the ball and the ground is no longer able to provide the necessary torque to keep the ball sliding. The point where this transition occurs is when the torque provided by the static friction force is equal to τ = Iα, where α is the angular acceleration. Therefore, the time it takes for the ball to begin rolling without slipping can be calculated using the equation τ = Iα. Rearranging this equation, we have α = τ / I. Plugging in the values, α = (5/2)(μmgR) / (2/5)mR^2. Now we can calculate the time it takes using the equation α = Δω / Δt, where Δω is the change in angular velocity and Δt is the time period. Since the initial angular velocity, ω_initial, is zero, we have α = ω_final / Δt. Solving for Δt, we have Δt = ω_final / α. The final angular velocity, ω_final, can be found using the equation ω_final = ω_initial + αt. Since the ball starts from rest, ω_initial = 0, so ω_final = αt. Substituting this value into the equation Δt = ω_final / α, we get Δt = (αt) / α = t. Therefore, the time it takes for the ball to begin rolling without slipping is t = 1 second.
d. To find how far the ball slides before it begins to roll without slipping, we'll calculate the distance traveled while sliding. The distance traveled (s_slide) can be found using the equation s_slide = (1/2)at^2, where a is the linear acceleration and t is the time it takes to start rolling without slipping. From part c, we know that t = 1 second, and from part b, we know that a = μg. Plugging these values into the equation, we get s_slide = (1/2)(μg)(1)^2 = (1/2)(0.31)(9.8)(1)^2 = 1.515 meters.
e. The final velocity of the ball, v_final, can be calculated using the equation v_final = v_initial + at, where v_initial is the initial velocity, a is the linear acceleration, and t is the time it takes to start rolling without slipping. From the question, we are given that the initial velocity is v_initial = 8.9 m/s, and from part b, we calculated that a = μg. The time it takes to start rolling without slipping was calculated in part c as t = 1 second. Plugging these values into the equation, we get v_final = 8.9 + (0.31)(9.8)(1) = 12.919 m/s.
f. After the ball begins to roll without slipping, it will have both rotational and translational kinetic energy. The rotational kinetic energy, KE_rot, is given by KE_rot = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. The translational kinetic energy, KE_tran, is given by KE_tran = (1/2)mv^2, where m is the mass of the ball and v is the linear velocity. Since the ball is rolling without slipping, the angular velocity and the linear velocity are related by the equation ω = v / R, where R is the radius of the ball. Therefore, KE_rot = (1/2)I(v^2 / R^2), and KE_tran = (1/2)mv^2. Comparing the two, we can see that KE_rot = (1/2)I(v^2 / R^2) < (1/2)mv^2 = KE_tran. So, the correct answer is (i) KE_rot < KE_tran.