- #1
Bonulo
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I've been presented with the following problems, and would like some help/affirmation:
SITUATION:
A billiard ball with mass m and radius r is in rest on a horizontal table. The ball is hit with a billiard cue the height r/3 above the table, and has the velocity v0 immediately after the hit.
The force from the cue F remains horizontal during the hit and the contact between cue and ball is short-termed. The kinetic coefficient of friction between the table and the ball is mu_k
PROBLEM a) Determine the ball's angular velocity omega0 immediately after the hit.
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Since the below-c.m. hit will give the ball backspin (draw English), the ball won't have natural roll. Therefore, I can't use the simple equation v0 = omega0*r, since that relationship only exist in natural roll.
Instead, I have to use Newton's 2nd law, sum_F = m*a and the rotational equivalent sum_tau = I * alpha, where tau is torque (around the c.m.), I is moment of inertia of the ball and alpha is its angular acceleration.
(1) sum_F = m*a <=> sum_F = m*alpha*r
<=> F - mu_k*m*g = m*alpha*r
(2) sum_tau = I*alpha <=> r*F - 2/3*r*mu_k*m*g = (2/3)*m*r^2*alpha
The two expressions each have only one unknown variable, alpha. In (1) alpha is the translational contribution to the angular acceleraton, and in (2) its the rotational contribution to the angular acceleration. Sum_alpha can now be determined. The hit has the duration dt, so omega0 = sum_alpha * dt. By substituting the expression for alpha into that equation I ought to have determined omega0. Right?
PROBLEM b) What's the velocity of the ball when it beings to do natural roll?
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The roll is natural when v = omega * r. Since the v0 and r is known, I should then be able to determine expressions for v and omega from the kinematic equations for the two types of motion. I guess the rolling friction can be ignored, since the needed coefficient isn't given. Therefore, there is no horizontal force. Right?
PROBLEM c) Determine the work of the friction force during the ball's movement until it starts rolling on the table.
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The ball starts at the angle 0 and reaches an angle theta. The torque is constant, since the kinetic friction is constant. Then the work done by the friction force must be W_f = tau*(theta-0) <=> W_f = tau*theta. Right?
:shy:
SITUATION:
A billiard ball with mass m and radius r is in rest on a horizontal table. The ball is hit with a billiard cue the height r/3 above the table, and has the velocity v0 immediately after the hit.
The force from the cue F remains horizontal during the hit and the contact between cue and ball is short-termed. The kinetic coefficient of friction between the table and the ball is mu_k
PROBLEM a) Determine the ball's angular velocity omega0 immediately after the hit.
-----
Since the below-c.m. hit will give the ball backspin (draw English), the ball won't have natural roll. Therefore, I can't use the simple equation v0 = omega0*r, since that relationship only exist in natural roll.
Instead, I have to use Newton's 2nd law, sum_F = m*a and the rotational equivalent sum_tau = I * alpha, where tau is torque (around the c.m.), I is moment of inertia of the ball and alpha is its angular acceleration.
(1) sum_F = m*a <=> sum_F = m*alpha*r
<=> F - mu_k*m*g = m*alpha*r
(2) sum_tau = I*alpha <=> r*F - 2/3*r*mu_k*m*g = (2/3)*m*r^2*alpha
The two expressions each have only one unknown variable, alpha. In (1) alpha is the translational contribution to the angular acceleraton, and in (2) its the rotational contribution to the angular acceleration. Sum_alpha can now be determined. The hit has the duration dt, so omega0 = sum_alpha * dt. By substituting the expression for alpha into that equation I ought to have determined omega0. Right?
PROBLEM b) What's the velocity of the ball when it beings to do natural roll?
----
The roll is natural when v = omega * r. Since the v0 and r is known, I should then be able to determine expressions for v and omega from the kinematic equations for the two types of motion. I guess the rolling friction can be ignored, since the needed coefficient isn't given. Therefore, there is no horizontal force. Right?
PROBLEM c) Determine the work of the friction force during the ball's movement until it starts rolling on the table.
----
The ball starts at the angle 0 and reaches an angle theta. The torque is constant, since the kinetic friction is constant. Then the work done by the friction force must be W_f = tau*(theta-0) <=> W_f = tau*theta. Right?
:shy: