1.

A solid cylinder of uniform density of radius 2 cm has mass of 50 g. If its length is 12 cm, calculate its moment of inertia about an axis passing through its centre and perpendicular to its length.

Answer» Given : R = 2 cm, M = 50 gm, L = 12 cm
Now, Moment of Inertia of cylinder, `I_("cm")=M((R^(2))/(4)+(L^(2))/(12))`
`=50xx10^(-3)[((2xx10^(-2))^(2))/(4)+((12xx10^(-2))^(2))/(12)]`
`=50xx10^(-3)xx[0.0001+0.0012]`
`=5xx10^(-2)xx13xx10^(-4)`
`=65xx10^(-6)`
`=6.50xx10^(-5)" kg m"^(2)`.


Discussion

No Comment Found

Related InterviewSolutions