What is the rotational kinetic energy of a baton of mass 0.5kg and length 0.6m which is rotating about an axis through its center at an angular speed of 3revs/sec ?
KE=1/2 I w^2
w= 3*2Pi rad/sec
I= 1/12 * .5*.6^2
To find the rotational kinetic energy of a baton, we can use the formula:
Rotational Kinetic Energy (KErot) = (1/2) * moment of inertia * angular velocity^2
Step 1: Find the moment of inertia (I) of the baton.
The moment of inertia depends on the shape and mass distribution of the object. For a baton rotating about an axis through its center, the moment of inertia can be calculated using the formula:
I = (1/12) * m * L^2
where m is the mass of the baton and L is the length of the baton.
Substituting the given values:
m = 0.5 kg
L = 0.6 m
I = (1/12) * 0.5 kg * (0.6 m)^2
Step 2: Calculate the rotational kinetic energy.
Now that we have the moment of inertia (I) and angular velocity (ω), we can substitute these values into the formula for rotational kinetic energy (KErot).
KErot = (1/2) * I * ω^2
Substituting the given values:
ω = 3 revs/sec
Note: Convert the angular velocity from revolutions per second to radians per second.
The conversion factor is 2π radians per revolution.
ω = 3 revs/sec * 2π radians/rev = 6π radians/sec
KErot = (1/2) * (1/12) * 0.5 kg * (0.6 m)^2 * (6π radians/sec)^2
Simplifying the expression will yield the final answer.
KErot = (1/2) * (1/12) * 0.5 kg * 0.36 m^2 * (36π^2) rad^2/sec^2
= 0.00125π^2 Joules or approximately 0.0123 Joules