A puck (mass m1 = 4.50 kg) slides on a frictionless table as shown in the figure

Question

A puck (mass m1 = 4.50 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 13.5 kg. The mass m2 is initially at a height of h = 13.5 m above the floor with the puck traveling in a circle of radius r = 3.60 m with a speed of 13.5 m/s. The force of gravity then causes mass m2 to move downward a distance 1.35 m. (a) What is the new speed of the puck? m/s (b) What is the change in the kinetic energy of the puck? J

Explanation / Answer

A puck (mass m1 = 4.30 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 12.9 kg. The mass m2 is initially at a height of h = 12.9 m above the floor with the puck traveling in a circle of radius r = 3.44 m with a speed of 12.9 m/s. The force of gravity then causes mass m2 to move downward a distance 1.29 m.

A.)What is the new speed of the puck?

B.)What is the new speed of the puck? A puck (mass m1 = 4.30 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 12.9 kg. The mass m2 is initially at a height of h = 12.9 m above the floor with the puck traveling in a circle of radius r = 3.44 m with a speed of 12.9 m/s. The force of gravity then causes mass m2 to move downward a distance 1.29 m.

A.)What is the new speed of the puck?

B.)What is the new speed of the puck? A puck (mass m1 = 4.30 kg) slides on a frictionless table as shown in the figure below. The puck is tied to a string that runs through a hole in the table and is attached to a mass m2 = 12.9 kg. The mass m2 is initially at a height of h = 12.9 m above the floor with the puck traveling in a circle of radius r = 3.44 m with a speed of 12.9 m/s. The force of gravity then causes mass m2 to move downward a distance 1.29 m.

A.)What is the new speed of the puck?

B.)What is the new speed of the puck? Best Answer Since the string is pulling to the center of the circle it will exert no torque on the circling puck. Since there is no torque angular momentum is constant. Lets calculate the angular momentum J before m2 moves:

J = moment of inertia x angular velocity
J = (m1r^2)(v/r) = m1vr = 4.3 x 12.9 x 3.44 = 190.82 kgm^2/s

When the mass m2 moves downward by 1.29m the radius of the circle will decrease by the same amount, giving a new radius = 3.44-1.29 = 2.15m
Thus, since ang momentum is conserved,

v' = J / m1r = 190.82 / (4.3)(2.15) = 20.64 m/s Best Answer Since the string is pulling to the center of the circle it will exert no torque on the circling puck. Since there is no torque angular momentum is constant. Lets calculate the angular momentum J before m2 moves:

J = moment of inertia x angular velocity
J = (m1r^2)(v/r) = m1vr = 4.3 x 12.9 x 3.44 = 190.82 kgm^2/s

When the mass m2 moves downward by 1.29m the radius of the circle will decrease by the same amount, giving a new radius = 3.44-1.29 = 2.15m
Thus, since ang momentum is conserved,

v' = J / m1r = 190.82 / (4.3)(2.15) = 20.64 m/s Since the string is pulling to the center of the circle it will exert no torque on the circling puck. Since there is no torque angular momentum is constant. Lets calculate the angular momentum J before m2 moves:

J = moment of inertia x angular velocity
J = (m1r^2)(v/r) = m1vr = 4.3 x 12.9 x 3.44 = 190.82 kgm^2/s

When the mass m2 moves downward by 1.29m the radius of the circle will decrease by the same amount, giving a new radius = 3.44-1.29 = 2.15m
Thus, since ang momentum is conserved,

v' = J / m1r = 190.82 / (4.3)(2.15) = 20.64 m/s

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