(Use the fact that the kinetic energy of a particle of mass "m" moving at speed "v" is (1/2)m(v^2). Slice objects into pieces in such a way that the velocity us approximately constant on each piece.)
Q: Find the kinetic energy of a rod of mass 10kg and length 6m rotating about an axis perpendicular to the rod at its midpoint, with an angular velocity of 2 radians per second. (Imagine a helicopter blade of uniform thickness.)
To find the kinetic energy of the rotating rod, we need to consider each infinitesimally small piece of the rod as a separate particle with its own mass and velocity. We can then calculate the kinetic energy of each piece and sum them up to get the total kinetic energy of the rod.
First, let's consider an infinitesimally small slice of the rod, located at a distance x from the midpoint. The mass of this slice can be determined by considering the linear density of the rod, which is the mass per unit length. In this case, the linear density is given by λ = m/L, where m is the mass of the rod and L is its length. Substituting the given values, we have λ = 10 kg / 6 m.
Now, we need to find the velocity of this slice. Since the rod is rotating with a constant angular velocity of 2 radians per second, the linear velocity of each slice is given by v = ω * x, where ω is the angular velocity and x is the distance of the slice from the axis of rotation. In this case, ω = 2 rad/s.
To calculate the kinetic energy of this infinitesimal slice, we can use the formula K.E. = (1/2) * m * v^2. In this case, the mass of the slice is dm = λ * dx, where dx is an infinitesimal length element.
Substituting the values, we have:
dm = λ * dx = (10 kg / 6 m) * dx
v = ω * x = 2 rad/s * x
Now we can calculate the kinetic energy of the slice:
dK.E. = (1/2) * dm * v^2 = (1/2) * (10 kg / 6 m) * dx * (2 rad/s * x)^2
To find the total kinetic energy of the rod, we integrate this expression over the length of the rod (from -L/2 to L/2) since the rod is rotating about its midpoint:
K.E. = ∫(dK.E.) = ∫[0, L/2] (1/2) * (10 kg / 6 m) * dx * (2 rad/s * x)^2 + ∫[-L/2, 0] (1/2) * (10 kg / 6 m) * dx * (2 rad/s * x)^2
Evaluating the integral, we get the kinetic energy of the rod rotating about an axis perpendicular to the rod at its midpoint.