A baggage carousel at an airport is rotating with an angular speed of 0.21 rad/s when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1100.0 kgm2. Ten pieces of baggage with an average mass of 24.0 kg each are dropped vertically onto the carousel and come to rest at a perpendicular distance of 2.00 m from the axis of rotation.
(a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed.
rad/s
(b) In reality, the angular speed of a baggage carousel does not change. Therefore, what must be the external torque acting on the system if ten bags are dropped each minute?
N m
An electric motor can accelerate a Ferris wheel of moment of inertia I = 21000 kg·m2 from rest to 10.4 rev/min in 12.0 s. When the motor is turned off, friction causes the wheel to slow down from 10.4 to 8.4 rev/min in 10.0 s.
(a) Determine the torque generated by the motor to bring the wheel to 10.4 rev/min.
N·m
(b) Determine the power that would be needed to maintain this rotational speed.
W
(a) To find the final angular speed of the baggage carousel, we can use the law of conservation of angular momentum. The initial angular momentum of the carousel can be calculated as the product of its moment of inertia and angular speed:
Initial angular momentum = moment of inertia * initial angular speed
L_initial = 1100.0 kgm^2 * 0.21 rad/s
Next, we can consider the angular momentum after the baggage is dropped onto the carousel. Since the bags come to rest at a perpendicular distance of 2.00 m from the axis of rotation, they contribute to the total angular momentum of the system. The angular momentum of each bag can be calculated as the product of its mass, perpendicular distance from the axis, and angular speed (which is zero since the bags come to rest):
Angular momentum of one bag = mass * distance * angular speed (which is 0)
L_bags = 10 bags * 24.0 kg/bag * 2.00 m * 0 rad/s
The total angular momentum of the system is conserved, so:
L_initial = L_bags
1100.0 kgm^2 * 0.21 rad/s = 10 bags * 24.0 kg/bag * 2.00 m * 0 rad/s
Now, solve for the final angular speed:
Final angular speed = (1100.0 kgm^2 * 0.21 rad/s) / (10 bags * 24.0 kg/bag * 2.00 m)
(b) In reality, the angular speed of a baggage carousel does not change. Therefore, to keep the angular speed constant while ten bags are dropped each minute, there must be an external torque acting on the system to counteract the change in angular momentum caused by the falling bags. The rate of change of angular momentum is given by:
Rate of change of angular momentum = (10 bags * 24.0 kg/bag * 2.00 m) / 60 s
To counteract this, an external torque must be applied in the opposite direction. Using the equation:
Torque = rate of change of angular momentum / change in time
Torque = ((10 bags * 24.0 kg/bag * 2.00 m) / 60 s) / 1 s
Now, calculate the torque:
Torque = (10 bags * 24.0 kg/bag * 2.00 m) / 60 s²
(a) To find the torque generated by the motor to bring the Ferris wheel to 10.4 rev/min, we can use the formula:
Torque = (change in angular momentum) / (change in time)
The change in angular momentum can be calculated as the product of the moment of inertia of the wheel and the change in angular speed:
Change in angular momentum = moment of inertia * (final angular speed - initial angular speed)
Change in angular momentum = 21000 kg·m² * (10.4 rev/min - 0 rev/min) / (1 min/60 s) * (2π rad/rev)
Next, calculate the change in time:
Change in time = 12.0 s
Now, calculate the torque:
Torque = (21000 kg·m² * (10.4 rev/min - 0 rev/min) / (1 min/60 s) * (2π rad/rev)) / 12.0 s
(b) To determine the power needed to maintain a rotational speed of 10.4 rev/min, we can use the formula:
Power = torque * angular speed
First, convert the angular speed to rad/s:
Angular speed = 10.4 rev/min * (2π rad/rev) / (1 min/60 s)
Now, calculate the power:
Power = (Torque from part (a)) * (Angular speed in rad/s)