Consider a physical pendulum constructed by attaching a thin, massless rod to the center of a uniform disc of mass 1.00 kg and radius 40.0 cm. The disc has a moment of inertia of ½ MR2 about an axis perpendicular to the plane of the disc, through its center. With the rod hung from one end (the pivot), the plane of the disc (attached to the other end) is vertical. If the rod has a length of 50.0 cm, then the period of (small amplitude) oscillations of the pendulum will be
Question
Consider a physical pendulum constructed by attaching a thin, massless rod to the center of a uniform disc of mass 1.00 kg and radius 40.0 cm. The disc has a moment of inertia of ½ MR2 about an axis perpendicular to the plane of the disc, through its center. With the rod hung from one end (the pivot), the plane of the disc (attached to the other end) is vertical. If the rod has a length of 50.0 cm, then the period of (small amplitude) oscillations of the pendulum will be
Solution
To solve this problem, we need to use the formula for the period of a physical pendulum:
T = 2π √(I / (mgh))
where:
- T is the period,
- I is the moment of inertia,
- m is the mass,
- g is the acceleration due to gravity, and
- h is the distance from the pivot point to the center of mass.
Given in the problem, we have:
- m = 1.00 kg,
- R = 40.0 cm = 0.4 m (we need to convert cm to m),
- L = 50.0 cm = 0.5 m (we need to convert cm to m),
- g = 9.81 m/s² (approximate value on Earth's surface).
The moment of inertia I for a disc rotating about an axis through its center and perpendicular to the plane of the disc is given by I = ½ MR². So, I = ½ * 1.00 kg * (0.4 m)² = 0.08 kg*m².
The distance h from the pivot point to the center of mass is the length of the rod plus the radius of the disc, h = L + R/2 = 0.5 m + 0.4 m / 2 = 0.7 m.
Substituting these values into the formula for the period, we get:
T = 2π √((0.08 kg*m²) / (1.00 kg * 9.81 m/s² * 0.7 m)) ≈ 1.07 s.
So, the period of small amplitude oscillations of the pendulum is approximately 1.07 seconds.
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