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A system consist of a ball of M2 and a uniform rod of M1 and length d. The rod is attached to a horizontal frictionless rotates at an angular speed w, as shown in the figure. The rotational inertia of the rod about point P is ? M1 d^2. The rod strikes the ball, which is initially at rest. As a result of this collision, the rod is stopped and the ball moves in the direction shown. Express all answers in terms of M1, M2, w, d, and fundamental constants.
A. Derive an expression from the angular momentum of the rod about point P before the collision.
B. Derive an expression for the speed v of the ball after the collision.
C. Assuming that this collision is elastic, calculate the numerical value of the ratio M1/M2
D. A new ball with the same mass M1 as the rod is now placed a distance x from the pivot as shown. Again assuming the collision is elastic, for what value x will the rod stop moving after hitting the ball?
A. To derive an expression for the angular momentum of the rod about point P before the collision, we need to consider the rotational inertia and angular velocity of the rod.
The rotational inertia of the rod about point P is given as M1d^2.
The angular momentum (L) of an object is defined as the product of its rotational inertia (I) and its angular velocity (ω). Therefore, the angular momentum of the rod about point P before the collision can be expressed as:
L = I * ω
Substituting the value of I as M1d^2 and ω as w, we have:
L = M1d^2 * w
Therefore, the expression for the angular momentum of the rod about point P before the collision is M1d^2w.
B. To derive an expression for the speed v of the ball after the collision, we can consider the conservation of angular momentum and kinetic energy.
Before the collision, the angular momentum of the system is solely due to the rod, which is M1d^2w.
After the collision, the rod stops rotating, so its angular velocity becomes zero. The entire angular momentum is then transferred to the ball.
The angular momentum of the ball can be expressed as:
L = I_ball * ω_ball
Since the ball is a point mass, its rotational inertia (I_ball) about point P is given as 0 (negligible).
Therefore, L_ball = 0 * ω_ball = 0
Since angular momentum is conserved, we can equate the initial angular momentum of the rod and the final angular momentum of the ball:
M1d^2w = 0
Solving for ω_ball, we get ω_ball = 0.
The linear speed of the ball can be calculated using the relation between linear speed (v) and angular speed (ω) of an object:
v = ω * r
where r is the distance between the pivot (point P) and the ball.
Since the rod rotates around P, the distance between P and the ball is d.
Therefore, the expression for the speed v of the ball after the collision is v = ω_ball * d = 0 * d = 0.
Hence, the ball comes to a complete stop after the collision.
C. To calculate the numerical value of the ratio M1/M2, we can use the conservation of kinetic energy during an elastic collision.
Before the collision, the rod has both rotational and translational kinetic energy. After the collision, the rod comes to a stop, so its kinetic energy becomes zero.
The kinetic energy of the rod before the collision can be expressed as:
KE_rod = (1/2) I_rod ω^2
where I_rod is the rotational inertia of the rod about point P (given as M1d^2) and ω is the initial angular velocity of the rod (given as w).
Substituting these values, we get:
KE_rod = (1/2) M1d^2 w^2
The kinetic energy of the ball after the collision can be expressed as:
KE_ball = (1/2) M2 v^2
Since the ball comes to a complete stop after the collision (as derived in part B), its kinetic energy becomes zero.
Hence, we can equate the initial kinetic energy of the rod to zero (as it comes to a stop) and the kinetic energy of the ball after the collision:
(1/2) M1d^2 w^2 = (1/2) M2 v^2
Simplifying and solving for M1/M2, we have:
M1/M2 = v^2 / (d^2w^2)
Since v is zero (as derived in part B), the ratio M1/M2 is also zero.
D. The new ball is placed a distance x from the pivot.
To find the value of x for which the rod stops moving after hitting the ball, we can again consider the conservation of total angular momentum of the system.
Before the collision, the angular momentum of the rod is M1d^2w.
After the collision, the rod stops rotating, so its angular velocity becomes zero.
The angular momentum of the ball can be expressed as:
L_ball = I_ball * ω_ball
Since the ball is a point mass, its rotational inertia (I_ball) about point P is given as M1d^2, as the ball has the same mass (M1) as the rod.
Therefore, L_ball = M1d^2 * ω_ball
Since angular momentum is conserved, we can equate the initial angular momentum of the rod and the final angular momentum of the ball:
M1d^2w = M1d^2 * ω_ball
Cancelling M1d^2 on both sides, we get:
w = ω_ball
For the rod to stop moving after hitting the ball, the final angular velocity (ω_ball) should be zero.
Therefore, we have:
w = 0
The angular velocity of the rod (w) is related to the linear speed (v) and distance (x) of the ball using the equation:
v = ω * x
Since w is zero, we need to find the value of x for which the linear speed v is also zero.
Hence, the rod will stop moving after hitting the ball when x = 0.
A. To derive an expression for the angular momentum of the rod about point P before the collision, we can use the formula for angular momentum:
Angular Momentum = moment of inertia * angular velocity
The moment of inertia of the rod about point P is given as M1 * d^2, and the angular velocity is w. Therefore, the angular momentum of the rod about point P before the collision is:
Angular Momentum = M1 * d^2 * w
B. To derive an expression for the speed v of the ball after the collision, we can use the principle of conservation of angular momentum. Since there is no external torque acting on the system, the total angular momentum before the collision should be equal to the total angular momentum after the collision.
Before the collision, the angular momentum is contributed only by the rod, as the ball is at rest. After the collision, the angular momentum is contributed by both the rod (which has stopped) and the ball, which is moving.
The angular momentum of the rod after the collision is zero, since it has stopped. The angular momentum of the ball after the collision can be expressed as:
Angular Momentum of ball = moment of inertia of ball * angular velocity of ball
The moment of inertia of the ball can be approximated as I ≈ 2/5 * M2 * r^2, where r is the radius of the ball.
Since the distance r is not given, we need to express it in terms of the given quantities. The ball gets its initial angular momentum from the rod, so we can equate the angular momentum of the ball to the angular momentum of the rod before the collision:
M1 * d^2 * w = 2/5 * M2 * r^2 * v
Solving for v, we get:
v = (5 * M1 * d^2 * w) / (2 * M2 * r^2)
C. Assuming the collision is elastic, we can use the principle of conservation of kinetic energy to find the ratio M1/M2. In an elastic collision, both momentum and kinetic energy are conserved.
Before the collision, the only object with kinetic energy is the rotating rod, as the ball is at rest. After the collision, the ball has taken on kinetic energy, while the rod has come to a stop.
The initial kinetic energy of the system is contributed by the rotating rod and can be expressed as:
Initial Kinetic Energy = (1/2) * M1 * (d^2) * (w^2)
The final kinetic energy of the system is contributed by the moving ball and can be expressed as:
Final Kinetic Energy = (1/2) * M2 * (v^2)
Since the collision is elastic, the initial kinetic energy should be equal to the final kinetic energy:
(1/2) * M1 * (d^2) * (w^2) = (1/2) * M2 * (v^2)
Simplifying and rearranging the equation, we can solve for the ratio M1/M2:
M1/M2 = (v^2 * (d^2))/(w^2)
D. To find the value of x for which the rod stops moving after hitting the ball, we can use the principle of conservation of angular momentum. Since the rod stops rotating, the total angular momentum of the system after the collision should be zero.
Before the collision, the angular momentum is given as M1 * d^2 * w. After the collision, the angular momentum is contributed by both the ball and the rod (which has stopped).
The angular momentum of the ball after the collision can be expressed as:
Angular Momentum of ball = moment of inertia of ball * angular velocity of ball
Since the ball starts from rest and the rod stops rotating, the angular momentum of the ball after the collision is:
Angular Momentum of ball = M2 * (x^2) * v
Setting the sum of the angular momenta equal to zero:
M1 * d^2 * w + M2 * (x^2) * v = 0
Substituting the expression for v derived in part B:
M1 * d^2 * w + M2 * (x^2) * [(5 * M1 * d^2 * w) / (2 * M2 * r^2)] = 0
Simplifying and solving for x:
x^2 = (-2 * M1 * d^2 * w) / (5 * r^2)
x = sqrt[(-2 * M1 * d^2 * w) / (5 * r^2)]
a) L=Iw
L=(1/3 md^2)w
b) L1=L2