The figure at the left shows a uniform disk of radius R = 0.8 m and 6-kg mass with as mall hole a distance d from the disk’s center that can serve as a pivot point. What should be the distance d so that the period of this physical pendulum is 2.7 s?
I didn't get any of that but thank you. I'll do my best to review it.
To determine the distance (d) from the disk's center to serve as the pivot point, we can use the equation for the period (T) of a physical pendulum.
The period of a physical pendulum is given by the formula:
T = 2 * π * √(I / (m * g * h))
Where:
T is the period of the pendulum,
π is a constant approximately equal to 3.14159,
I is the moment of inertia of the pendulum about the pivot point,
m is the mass of the pendulum,
g is the acceleration due to gravity, and
h is the distance from the pivot point to the center of mass of the pendulum.
The moment of inertia (I) of a uniform disk rotating about an axis perpendicular to its plane and passing through its center is given by the formula:
I = (1/2) * m * R^2
Where:
m is the mass of the disk, and
R is the radius of the disk.
Substituting the formula for the moment of inertia into the formula for the period, we get:
T = 2 * π * √(((1/2) * m * R^2) / (m * g * h))
Simplifying further:
T = 2 * π * √((R^2) / (4 * g * h))
To find the distance (d) from the disk's center that serves as the pivot point, we need to rearrange the equation:
d = R - h
Squaring both sides of the equation:
d^2 = (R - h)^2
Rearranging:
h = R - √(d^2)
Now we can substitute this value of h back into the equation for the period:
T = 2 * π * √((R^2) / (4 * g * (R - √(d^2))))
Finally, we can solve this equation for the given period (T = 2.7 s) and the known values of R (0.8 m), m (6 kg), and g (9.8 m/s^2) to find the distance d.
I around center
Assume you do not have a moment of inertia table so
integral rho (2 pi r t) r^2 dr = rho 2 pi t integral r^3 dr
= rho 2 pi t R^4/4 = rho pi t R^4/2
mass = rho pi t R^2
so integral= mass * R^2/2 = I around center
I around point distance d from center =m d^2 + m R^2/2
at angle T radians from straight down
gravity perpendicular to line to center = m g d sinT
for small T that is m g d T
torque about center = -m g d T
torque = -I d^2T/dt^2
-m g d T = [ m d^2 + m R^2/2 ] d^2 T/ dt^2
if T = a sin w t
d^2 T/dt^2 = - a w^2sin wt = - w^2 T
so
-m g d T = - [ m d^2 + m R^2/2 ] w^2 T
w^2 = g d /[ d^2 + R^2/2]
w = 2 pi f = 2 pi /period
4 pi^2 / period = g d /[ d^2 + R^2/2]