A toy consisting of two balls, each m = 0.56 kg, at the ends of a l = 0.73 m long, thin light-weight rod is shown below. Find the moment of inertia of the toy. The moment of inertia is to be found about the center of the rod.
To find the moment of inertia of the toy about the center of the rod, we need to use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of an object about an axis parallel to and a distance "d" away from the center of mass is equal to the moment of inertia about the center of mass plus the mass of the object times the square of the distance "d".
In this case, the moment of inertia of the toy about the center of the rod can be found by summing the moment of inertia of each ball separately.
The moment of inertia of a ball rotating about its own center of mass is given by the formula:
I = (2/5) * m * r^2
where m is the mass of the ball and r is the radius of the ball.
Since the rod is thin and lightweight, we can assume its mass is negligible compared to the balls. Therefore, its moment of inertia can be ignored.
Substituting the values into the formula, we have:
I1 = (2/5) * m * r^2
I2 = (2/5) * m * r^2
where I1 and I2 are the moment of inertia of each ball, and r is the distance from the center of the rod to each ball. In this case, since the rod is l = 0.73 m long, the distance "r" will be half of the rod's length: r = l/2 = 0.73/2 = 0.365 m.
Substituting the known values into the formula, we get:
I1 = (2/5) * 0.56 kg * (0.365 m)^2
I2 = (2/5) * 0.56 kg * (0.365 m)^2
Finally, to find the total moment of inertia of the toy about the center of the rod, we add the moment of inertia of each ball together:
πΌπ‘ππ‘ππ = πΌ1 + πΌ2
= (2/5) * 0.56 kg * (0.365 m)^2 + (2/5) * 0.56 kg * (0.365 m)^2
= 2 * (2/5) * 0.56 kg * (0.365 m)^2
= 0.12864 kg * (0.365 m)^2
= 0.0173177608 kg*m^2
So, the moment of inertia of the toy about the center of the rod is approximately 0.0173 kg*m^2.