If a trapeze artist rotates once each second while sailing through the air and contracts to reduce her rotational inertia to one-third of what it was, how many rotations per second will result?
3 rotations/sec, since the
(moment of inertia)x(rotation rate) product must stay the same. It is a measure of angular momentum.
3 rotations/sec, since the
(moment of inertia)x(rotation rate) product must stay the same. It is a measure of angular momentum.
To find the number of rotations per second after the trapeze artist contracts, we need to consider the conservation of angular momentum.
Angular momentum is given by the equation L = Iω, where:
L = angular momentum
I = rotational inertia
ω = angular velocity
Initially, the angular momentum of the trapeze artist is conserved:
L1 = I1ω1
After contracting, the rotational inertia becomes one-third of the initial value:
I2 = (1/3)I1
The angular momentum remains conserved:
L2 = I2ω2
Since the angular momentum is conserved, L1 = L2:
I1ω1 = I2ω2
Substituting the values we know:
I1ω1 = (1/3)I1ω2
Simplifying the equation by canceling out I1 and rearranging:
ω2 = 3ω1
Since initially the trapeze artist rotates once per second (ω1 = 1), we can calculate the new angular velocity (ω2):
ω2 = 3 × 1 = 3 rotations per second
Therefore, after contracting and reducing the rotational inertia to one-third, the trapeze artist will rotate at a rate of 3 rotations per second.
To determine the number of rotations per second after the trapeze artist contracts, we first need to understand the concept of rotational inertia or moment of inertia.
Rotational inertia is a property of an object that measures its resistance to changes in rotational motion. It depends on the mass distribution and the axis of rotation. The rotational inertia of an object is given by the formula:
I = m * r^2
Where:
- I is the rotational inertia
- m is the mass of the object
- r is the distance between the object and the axis of rotation.
In the given scenario, the trapeze artist is rotating once each second initially. Let's assume her initial rotational inertia is I_initial.
Now, when she contracts to reduce her rotational inertia to one-third of the initial value, we can express her new rotational inertia as:
I_final = (1/3) * I_initial
Given that the trapeze artist is still rotating at the same speed (once per second), we can equate the rotational inertia equations:
I_initial = I_final
m * r^2 = (1/3) * m * r^2
We can see that the mass term cancels out, giving us:
r^2 = (1/3) * r^2
Simplifying the equation further, we have:
1 = 1/3
This equation shows that our assumption for the rotational inertia reduction is incorrect. It is not possible to reduce the rotational inertia to one-third of the initial value while still rotating at the same speed (once per second). Therefore, we cannot determine the number of rotations per second that will result from this scenario.