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

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: