Question Part
Points
Submissions Used
A thin cylindrical rod of length 3.00 m and mass 1.5 kg has two spheres attached on its ends. The centers of the spheres are 1.50 m from the center of the rod. The mass of each sphere is M = 0.72 kg. The rod and spheres are initially at rest on a frictionless icy surface; the figure below is drawn looking down on that icy surface, with north towards the top of the page. A small mass m = 0.23 kg, sliding across the frictionless ice with a speed of
v = 1.5 m/s
and in the Eastward direction as shown, strikes the southernmost sphere and sticks to it. At the same instant, an identical mass having twice the speed, and also traveling to the East, strikes the northernmost sphere and also sticks. As a result of the simultaneous collisions, the entire system starts to translate and also to rotate.
(a) Determine the speed of the center of mass of the rod after the simultaneous collisions.
m/s
(b) Determine the magnitude and direction of the angular velocity of the system after the simultaneous collisions. You may treat the two moving masses and the two spheres at the ends of the rod as if they were point particles.
magnitude rad/s
direction
To solve this problem, we can apply the principle of conservation of linear momentum and the principle of conservation of angular momentum.
(a) To determine the speed of the center of mass of the rod after the simultaneous collisions, we can use the principle of conservation of linear momentum. The linear momentum before the collision is equal to the linear momentum after the collision.
The total initial linear momentum of the system is given by:
P_initial = m*v + m*(2v) = 0.23 kg * 1.5 m/s + 0.23 kg * (2 * 1.5 m/s)
Here, we have two identical masses, each with a speed of 1.5 m/s, and another identical mass with twice the speed. So, the total initial linear momentum is:
P_initial = 0.23 kg * 1.5 m/s + 0.23 kg * (2 * 1.5 m/s) = 1.725 kg·m/s + 1.725 kg·m/s = 3.45 kg·m/s
The total final linear momentum of the system will be equal to the mass of the whole system (rod + spheres + masses) times the final velocity of the center of mass. Let's denote the final velocity of the center of mass as V_cm.
P_final = (m_rod + 2m_sphere + 2m_mass) * V_cm
The mass of the rod, m_rod, is given as 1.5 kg. The mass of each sphere, m_sphere, is given as 0.72 kg. And the mass of each mass, m_mass, is given as 0.23 kg.
P_final = (1.5 kg + 2 * 0.72 kg + 2 * 0.23 kg) * V_cm
To solve for V_cm, we can equate the initial and final linear momenta and solve for V_cm:
P_initial = P_final
3.45 kg·m/s = (1.5 kg + 2 * 0.72 kg + 2 * 0.23 kg) * V_cm
Simplifying, we get:
V_cm = 3.45 kg·m/s / (1.5 kg + 2 * 0.72 kg + 2 * 0.23 kg)
Calculating the expression on the right-hand side will give us the speed of the center of mass of the rod after the collisions.
(b) To determine the magnitude and direction of the angular velocity of the system after the simultaneous collisions, we can use the principle of conservation of angular momentum. The angular momentum before the collision is equal to the angular momentum after the collision.
The initial angular momentum of the system is zero since the rod and spheres are initially at rest. Therefore, the final angular momentum of the system is also zero.
The angular momentum of a point particle is given by the equation:
L = I * ω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since the rod and the spheres attached to its ends are thin and cylindrical, we can treat them as point particles. The moment of inertia of a point particle is given by the equation:
I = m * r^2
Where m is the mass of the point particle and r is the distance of the point particle from the axis of rotation.
Let's denote the distance of the spheres from the center of the rod as d. According to the given information, d = 1.5 m.
The moment of inertia of each sphere is:
I_sphere = m_sphere * d^2
Since we have two identical spheres, the total moment of inertia for the two spheres is:
I_total = 2 * I_sphere
The moment of inertia of the rod can be calculated using the equation for a thin cylinder:
I_rod = (1/12) * m_rod * L_rod^2
Where L_rod is the length of the rod. According to the given information, L_rod = 3.00 m.
Now, the total moment of inertia for the whole system is the sum of the moment of inertia of the rod and the moment of inertia of the spheres:
I_total = I_rod + I_total
To find the angular velocity, ω, we can equate the initial and final angular momenta and solve for ω:
L_initial = L_final
0 = I_total * ω
Simplifying, we get:
ω = 0 / I_total
Calculating the expression on the right-hand side will give us the magnitude of the angular velocity of the system after the collisions.
To determine the direction of the angular velocity, we need to consider the orientation of the system after the collisions. Since the block collides with the spheres and sticks to them, the system will rotate counterclockwise when viewed from above. Therefore, the direction of the angular velocity is counterclockwise.
Now, let's compute the values for the speed of the center of mass of the rod and the magnitude of the angular velocity.