A spherical bowling ball with mass m = 3.2 kg and radius R = 0.107 m is thrown down the lane with an initial speed of v = 8.6 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.34. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.
1)What is the magnitude of the angular acceleration of the bowling ball as it slides down the lane?
77.85Rad/sec^2
2)What is magnitude of the linear acceleration of the bowling ball as it slides down the lane?
3.332m/s^2
3)How long does it take the bowling ball to begin rolling without slipping?
.737437sec
4)How far does the bowling ball slide before it begins to roll without slipping?
5.43587
5)What is the magnitude of the final velocity?
I figured out the whole problem except for 5, i can't seem to get the right answer. Does anyone know it?
To find the magnitude of the final velocity of the bowling ball, we need to consider the transition from sliding to rolling without slipping. When the ball is sliding, it is subject to kinetic friction, which opposes its motion. Once the ball starts rolling without slipping, the friction force changes from kinetic friction to static friction, which provides the necessary torque to produce angular acceleration.
Let's break down the problem step by step:
1) To find the angular acceleration (α) of the bowling ball as it slides down the lane, we can use the following equation:
τ = Iα
where τ is the torque applied to the ball, I is the moment of inertia, and α is the angular acceleration. Since the ball is a solid sphere, the moment of inertia can be calculated using the formula:
I = (2/5) * m * R^2
Substituting the given values, we find:
I = (2/5) * 3.2 kg * (0.107 m)^2 = 0.02567 kg*m^2
Next, we need to calculate the torque applied to the ball. The torque created by the friction force (τ_friction) can be found using the equation:
τ_friction = μ * R * F_n
where μ is the coefficient of friction, R is the radius of the ball, and F_n is the normal force. The normal force can be calculated using the equation F_n = mg, where g is the acceleration due to gravity.
F_n = 3.2 kg * 9.8 m/s^2 = 31.36 N
Substituting the values, we have:
τ_friction = 0.34 * 0.107 m * 31.36 N = 1.1345 N*m
Since the friction force opposes the motion, the torque is negative. Hence, τ = -1.1345 N*m.
Plugging the values into the equation τ = Iα, we can solve for α:
-1.1345 N*m = 0.02567 kg*m^2 * α
α = (-1.1345 N*m) / (0.02567 kg*m^2) = -44.213 rad/s^2
The negative sign indicates that the ball experiences angular deceleration.
The magnitude of the angular acceleration is the absolute value of α, so |α| = 44.213 rad/s^2. Therefore, the magnitude of the angular acceleration of the bowling ball as it slides down the lane is approximately 44.213 rad/s^2.
2) The linear acceleration (a) of the bowling ball as it slides down the lane is given by:
a = R * α
Substituting the values, we get:
a = 0.107 m * (-44.213 rad/s^2) = -4.732 m/s^2
The negative sign indicates that the ball experiences linear deceleration.
The magnitude of the linear acceleration is the absolute value of a, so |a| = 4.732 m/s^2. Therefore, the magnitude of the linear acceleration of the bowling ball as it slides down the lane is approximately 4.732 m/s^2.
3) To find the time it takes for the bowling ball to begin rolling without slipping, we can use the formula:
t_roll = v / a
where v is the initial velocity and a is the linear acceleration. Substituting the given values:
t_roll = 8.6 m/s / 4.732 m/s^2 = 1.820 s
Therefore, it takes approximately 1.820 seconds for the bowling ball to begin rolling without slipping.
4) To determine the distance the bowling ball slides before it begins to roll without slipping, we can use the equation:
s_slide = (1/2) * a * t_roll^2
Substituting the values, we get:
s_slide = (1/2) * 4.732 m/s^2 * (1.820 s)^2 = 7.502 m
Therefore, the bowling ball slides approximately 7.502 meters before it begins to roll without slipping.
5) Finally, we need to find the magnitude of the final velocity (v_final) of the bowling ball. Since the ball rolls without slipping, the linear velocity at the point of transition is equal to the distance traveled (s_slide) divided by the time taken (t_roll). Therefore, we have:
v_final = s_slide / t_roll
Substituting the values, we can calculate:
v_final = 7.502 m / 1.820 s = 4.119 m/s
Therefore, the magnitude of the final velocity of the bowling ball is approximately 4.119 m/s.