Calculate the moment of incrtia of a uniform hollow sphere of negligible thickness of mass m and

Calculate the moment of incrtia of a uniform hollow sphere...

Key Concepts

Moment of Inertia

This is a measure of an object's resistance to angular acceleration when subjected to a torque, depending not only on the object's mass but also on how the mass is distributed relative to the axis of rotation.

Uniform Mass Distribution

This concept indicates that the object's mass is distributed evenly throughout, which simplifies the calculation of physical quantities such as moment of inertia since the density remains constant over the object's surface or volume.

Thin Shell (Hollow Sphere) Approximation

In the thin shell approximation, the thickness of the sphere is negligible compared to its radius, meaning the entire mass is considered to be concentrated at a fixed distance from the center. This allows for the use of simplified formulas when calculating properties like the moment of inertia.

Axis of Rotation

The moment of inertia depends on the choice of the axis about which the rotation occurs. For objects with symmetrical shapes, such as spheres, the axis through the center (like a diameter) is commonly used, and its symmetry assists in the straightforward computation of rotational inertia.

Transcript

00:02 Hello everyone, we are going to understand this question here given in the question there is a hollow sphere given there is a hollow sphere having mass m, mass is equal to m and radius that is given r we have to find the movement of inertia about the diameter this is the x -axis which is representing the diameter so surface mass density of holo -spier surface mass density that is represented by sigma which is equal to total mass upon total surface area that is 4 pi r square so we can write taking the partial area that is represented by d a which is a part of circular strip or ring so we can write d a is equal to 2 pi into r into this curved part this is a circular ring type, partial hollow sphere.

01:37 So its surface area is 2 pi 2 pi r into 2 pi r squared sine theta into t theta.

01:58 So partial mass density dm is equal to sigma into d a.

02:05 Sigma into d a or we can write dm is equal to m upon 4 pi r square into 2 pi r square sine theta into d theta so here 2 pi r square and 2 pi r square will get cancelled so after calculation we will get m by 2 into sine theta into d theta sine theta into d theta so moment of inertia now moment of inertia d i is equal to d m into r square because it is in the form of circular ring now substituting the value in this here value of dm is m upon 2 into sine theta into d theta into r square now we can write here value of r r square sine theta into d theta sine theta square so here it is sine square theta so we can write m into r square upon 2 into sine cube theta into d theta now taking the integration of both side now taking integration taking integration so we can write i is equal to integration of d -i or we can write m r square upon 2 into integration of sine cube theta into d -theta and integration limit is from 0 to r again we can write mr square upon 2 into sine square theta into sine theta into sine theta into d theta sine square theta into sine theta into sine theta into d theta now we can write if we are taking t is equal to cost theta then d t is equal to d t is equal to sine theta sorry minus sine theta into d theta and we know that sine square theta is equal to 1 minus cos square theta so substituting this value here we will get i is equal to m r square upon 2 and integration limit is from 0 to r here sine square theta is equal to 1 minus cos square theta into sine theta into d theta now we have supposed that t is equal to cost theta now substituting the value in this here we can write m r square upon 2 into theta is from 0 to sorry here here theta is from 0 to pi by mistakenly i have retained it at 0 to r this is integration limit is from 0 to 0 to let's correct it here this is integration limit because here we are integrating with respect to theta so here theta is from 0 to pi 0 to now here we can write we can write 1 minus t square into d t and this is minus because we have already calculated d t is minus sin theta into d theta.

07:25 So here we have just substituted the values.

07:29 Now taking the integration with respect to t.

07:35 So we will get mr square upon 2 and here integration is from we will get minus t plus t cube upon 3 and integration limit is from 0 to pi...

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