Each wheel is a solid disk of mass 2.0 kg. On the horizontal surfaces the center of mass of each wheel moves with a linear speed of 6.0 m/s.
(a1) What is the algebraic expression for the total kinetic energy KEtotal of the rolling wheel? Express your answer in terms of the mass m of the wheel and the linear speed v of the center of mass of the wheel.
KEtotal =
How can I find the expression?
(1/2) m v^2 + (1/2) I omega^2
but
omega = v/R
so
(1/2) m v^2 + (1/2) I v^2/R^2
but I
I = (1/2) m R^2
so
(1/2) m v^2 + (1/2)(1/2) m R^2 v^2/R^2
etc
To find the expression for the total kinetic energy (KEtotal) of the rolling wheel, you need to consider two types of kinetic energy: translational kinetic energy and rotational kinetic energy.
Translational kinetic energy (KEtrans) is given by the equation:
KEtrans = (1/2) * m * v^2
where m is the mass of the wheel and v is the linear speed of the center of mass of the wheel.
Rotational kinetic energy (KErot) is given by the equation:
KErot = (1/2) * I * ω^2
where I is the moment of inertia of the wheel and ω is the angular velocity of the wheel.
For a solid disk rotating about its center, the moment of inertia (I) can be calculated using the formula:
I = (1/2) * m * r^2
where r is the radius of the wheel.
Since the linear speed (v) of the center of mass of the wheel is related to the angular velocity (ω) by the equation:
v = r * ω
you can rewrite the rotational kinetic energy equation as:
KErot = (1/2) * I * (v/r)^2
Substituting the moment of inertia (I) with its value, you get:
KErot = (1/2) * (1/2) * m * r^2 * (v/r)^2
Simplifying the expression, you have:
KErot = (1/4) * m * v^2
To find the total kinetic energy (KEtotal), you simply add the translational and rotational kinetic energies together:
KEtotal = KEtrans + KErot
= (1/2) * m * v^2 + (1/4) * m * v^2
= (3/4) * m * v^2
Therefore, the algebraic expression for the total kinetic energy (KEtotal) of the rolling wheel is:
KEtotal = (3/4) * m * v^2