We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What must the hoop’s radius be?

Explanation:

Given that the hoop makes one complete oscillation in 2s, this implies that the period of oscillation is 2s

Then,

T = 2s

Let Mass of the thin hoop be M

Let Radius of the hoop be R

Moment of inertial of a hoop is

I = MR²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

M is the mass of the hoop

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = R

Then, applying the formula

T = 2π √ (I / MgR)

So, we know that

I = MR²

Then,

T = 2π √( MR² / MgR)

T = 2π √(R/g)

Make R subject of formula

Square both sides

T² = 4π²(R/g)

T²g = 4π²R

Then, R = T²g / 4 π²

Since g = 9.8m/s² and T= 2s

R = 2² × 9.8 / 4π²

R = 0.993m

Then, the radius of the hoop is 0.993m

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

explanation:

prndulum needs thread, weight clock to time swings