In each of the following cases, the mass is 7.99 kg and the radius is 7.50 cm.
(a) What is the moment of inertia (kg/m2
) of a solid cylinder rotating about an axis parallel to the symmetry axis but passing through the edge of the cylinder?
(b) What is the moment of inertia (kg/m2
) of a solid sphere rotating about an axis tangent to its surface?
The wording in (a) is confusing me. I tried but wasn't even close on both.
a)=0.0674
b)=0.0629
here are the answers can someone help me figure out how?
Read and apply the parallel axis theorem.
http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
To calculate the moment of inertia for each case, we need to use the appropriate formulas. Let's break down the problem step-by-step:
(a) Solid Cylinder Rotating about an Axis Passing through the Edge:
In this case, the moment of inertia can be calculated using the formula:
I = 0.5 * m * r^2
where
I is the moment of inertia,
m is the mass of the object, and
r is the radius from the axis of rotation.
Plugging in the given values:
m = 7.99 kg
r = 7.50 cm = 0.075 m
Calculating:
I = 0.5 * (7.99 kg) * (0.075 m)^2
I = 0.5 * 7.99 * 0.005625
I = 0.02803 kg/m^2 (approximately)
So, the moment of inertia for a solid cylinder rotating about an axis passing through the edge is approximately 0.02803 kg/m^2.
(b) Solid Sphere Rotating about an Axis Tangent to Its Surface:
The moment of inertia for a solid sphere rotating about an axis tangent to its surface can be calculated using the formula:
I = 0.4 * m * r^2
Plugging in the given values:
m = 7.99 kg
r = 7.50 cm = 0.075 m
Calculating:
I = 0.4 * (7.99 kg) * (0.075 m)^2
I = 0.4 * 7.99 * 0.005625
I = 0.11998 kg/m^2 (approximately)
So, the moment of inertia for a solid sphere rotating about an axis tangent to its surface is approximately 0.11998 kg/m^2.
I hope this helps clarify how to calculate the moment of inertia in each case!