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A pendulum is constructed from a heavy metal rod and a metal disk, both of uniform mass density. The center of the disk is bolted to one end of the rod, and the pendulum hangs from the other end of the rod. The rod has a mass of =1.0 kg and a length of =49.8 cm. The disk has a mass of =4.0 kg and a radius of =24.9 cm. The acceleration due to gravity is =9.8 m/s2.
The pendulum is held with the rod horizontal and then released. What is the magnitude of its angular acceleration at the moment of release?

Respuesta :

The magnitude of the angular acceleration of the pendulum at the moment of release is; α = 18.45 rad/s²

We are given;

Mass of rod; m = 1 kg

Length of rod; L = 49.8 cm = 0.498 m

Mass of Disk; M = 4 kg

Radius of disk; r = 24.9 cm = 0.249 m

Let us first calculate the torque acting from the formula;

τ = mg(L/2) + MgL

Thus;

τ = (1 × 9.8 × (0.498/2)) + (4 × 9.8 × 0.498)

τ = 21.96 N.m

Using parallel axis theorem, we can find the moment of inertia about the given axis as;

I = (mL²/3) + ½MR² + ML²

Plugging in the relevant values gives;

I = (1 * 0.498²/3) + ½(4 * 0.249²) + (4 * 0.498²)

I = 1.19 kg.m²

The angular acceleration is given by the formula;

α = I/τ

α = 21.96/1.19

α = 18.45 rad/s²

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