Question

A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as indicated below. Its center of mass is a distance d=0.300 m from the pivot axis and its moment of inertia about an axis through the CM and parallel to the pivot axis is ICM=9.00×10-3 kg m2.
The plate is released from an angle θ=15°. Calculate the period of the oscillations.

Answer 1

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2$\pi \sqrt{0.5*9.8*0.3}$

T=1.2042s

The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

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