A 90 kg boy jogs on a circular track with a radius of 40 m at 5 m / s. What is the boy’s angular momentum with respect to the center of the track?
L = I omega
I = m r^2 = 90 * 40^2
omega = v/r = 5 /40
so
L = 90 * 40^2 * 5 /40
= 90 * 200
= 18000
To find the boy's angular momentum with respect to the center of the track, we can use the formula for angular momentum:
Angular momentum (L) = moment of inertia (I) * angular velocity (ω)
where moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion, and angular velocity (ω) is the rate at which an object rotates.
The moment of inertia of a point mass rotating about a fixed axis at a distance r is given by:
Moment of inertia (I) = mass (m) * radius squared (r^2)
In this case, the boy is the point mass, the radius of the track is 40 m, and the mass of the boy is 90 kg. So we can calculate the moment of inertia:
I = 90 kg * (40 m)^2 = 144,000 kg.m^2
The angular velocity (ω) is given as 5 m/s, which is the linear velocity. To find the angular velocity, we can use the formula:
Angular velocity (ω) = linear velocity (v) / radius (r)
ω = 5 m/s / 40 m = 0.125 rad/s
Now we can substitute the values into the formula for angular momentum:
L = I * ω = 144,000 kg.m^2 * 0.125 rad/s = 18,000 kg.m^2/s
Therefore, the boy's angular momentum with respect to the center of the track is 18,000 kg.m^2/s.