In the experiment, you will study a torsion pendulum and compare the periods of a suspended sphere and of a suspended cube. The rotational inertia of a sphere is Is = 1/10(MsD^2), where Ms is the mass of the sphere and D is its diameter. The rotational inertia of a cube is Ic = 1/6(McS^2), where Mc is the mass of the cube and S is the length of its side. If the cube and the sphere are suspended from the same wire, what is the expected ratio of their periods, Tc/Ts?
Assume that D = S,
Ms = 0.20kg,
and Mc = 0.6 kg.
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To find the ratio Tc/Ts, I tried doing:
(1/6 * 0.6) / (1/10 * .2) = 5, but this is incorrect.
What am I missing?
To find the expected ratio of the periods Tc/Ts, there are a few steps you need to follow correctly.
1. Calculate the rotational inertia of the sphere (Is) using the given formula:
Is = (1/10) * (Ms * D^2)
2. Calculate the rotational inertia of the cube (Ic) using the given formula:
Ic = (1/6) * (Mc * S^2)
3. Simplify the formulas of rotational inertia using the fact that D = S:
Is = (1/10) * (Ms * D^2) = (1/10) * (Ms * S^2) (since D = S)
Ic = (1/6) * (Mc * S^2)
4. Calculate the ratio of the periods Tc/Ts using the formula:
Tc/Ts = sqrt(Ic/Is)
Now, let's calculate the ratio of the periods Tc/Ts step by step.
Step 1:
Is = (1/10) * (0.20 * S^2)
Is = 0.02 * S^2
Step 2:
Ic = (1/6) * (0.6 * S^2)
Ic = 0.1 * S^2
Step 3:
Tc/Ts = sqrt(Ic/Is)
Tc/Ts = sqrt((0.1 * S^2) / (0.02 * S^2))
Tc/Ts = sqrt(5)
Therefore, the expected ratio of their periods Tc/Ts is sqrt(5), which is approximately equal to 2.236.
To find the expected ratio of the periods Tc/Ts, we need to consider the formula for the period of a torsion pendulum. The period of a torsion pendulum is given by T = 2π√(I/k), where I is the rotational inertia and k is the torsion constant.
First, let's find the rotational inertia of the sphere (Is) and the rotational inertia of the cube (Ic) using the given formulas:
Is = (1/10) * (Ms * D^2)
= (1/10) * (0.20 kg * (D * D))
Ic = (1/6) * (Mc * S^2)
= (1/6) * (0.6 kg * (D * D))
We are given that D = S, so we can substitute S with D in the formula for the cube's rotational inertia (Ic):
Ic = (1/6) * (0.6 kg * (D * D))
= (1/6) * (0.6 kg * (D^2))
Now, let's calculate the expected ratio of the periods Tc/Ts:
Tc/Ts = (2π√(Ic/k)) / (2π√(Is/k))
= √((Ic/k) / (Is/k))
Since both the sphere and the cube are suspended from the same wire, the torsion constant (k) is the same for both. Therefore, we can cancel out the k terms in the ratio:
Tc/Ts = √(Ic / Is)
= √(((1/6) * (0.6 kg * (D^2))) / ((1/10) * (0.20 kg * (D^2))))
Now, substitute D with the given value:
Tc/Ts = √(((1/6) * (0.6 kg * (D^2))) / ((1/10) * (0.20 kg * (D^2))))
= √(((1/6) * (0.6 kg * (D^2))) / ((1/10) * (0.20 kg * (D^2))))
Simplifying the expression:
Tc/Ts = √((1/6) * 3)
= √(1/2)
= (√1) / (√2)
= 1 / (√2)
≈ 0.707
Therefore, the expected ratio of the periods Tc/Ts is approximately 0.707.