Question

moment of inertia of a uniform horizontal solid cylinder of mass M about an Axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius R is

Answer 1

your answer is : I = M (13R^2/4).

Am I Right????

Answer 2

Answer:

Moment of inertia is $49 \mathrm{MR}^2 / 4$

Explanation:

Explanation:Step 1: A (solid) cylinder is a solid that has two parallel surfaces and a cylindrical surface surrounding it. An element of the cylinder is a line segment that is decided by an element of the cylindrical surface between two parallel planes.

A cylinder has components with equal lengths.The moment of inertia of a solid cylinder about its own axis is the same as its moment of inertia about an axis passing through its point of gravity and perpendicular to its length. Its length L and radius R have the relationship: L=2R.

Step 2: Rolling along a horizontal surface without slipping is possible with a solid cylinder connected to a horizontal massless spring.

A cylinder is a three-dimensional solid figure with two sides that are identically circular and are connected by a curved surface that is located at a specific height from the centre. Real-world instances of cylinders include toilet paper rolls and cold beverage cans.

From parallel axix theorem

$\mathrm{I}=\mathrm{M}\left[\frac{\mathrm{L}^2}{12}+\frac{\mathrm{L}^2}{4}\right]+\mathrm{M}\left[\frac{\mathrm{R}}{2}\right]^2$

$\mathrm{I}=\frac{\mathrm{ML}^2}{3}+\frac{\mathrm{MR}^2}{4}$

$\text { given } L=6 R$$\mathrm{L}=\frac{\mathrm{M}}{3}(6 \mathrm{R})^2+\frac{\mathrm{MR}^2}{4}=\frac{49}{4} \mathrm{MR}^2$

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