A hollow sphere of radius 0.15 m, with rotational inertia I=0.040 kg ·m^2 about a line through

A hollow sphere of radius 0.15 m, with rotational inertia...

Key Concepts

Rolling Motion Dynamics

Rolling motion involves an object that rotates about its own axis while translating along a surface. In this type of motion, the movement of the object is a combination of translation (the motion of its center of mass) and rotation (the object spinning about its center of mass), which must be analyzed together, especially when the object rolls without slipping.

Kinetic Energy Decomposition in Rolling Objects

The total kinetic energy of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy. Translational energy is due to the motion of the center of mass, expressed as (1/2)mv², while rotational energy is due to the rotation about the center of mass, expressed as (1/2)I?². This decomposition is essential for understanding energy distribution in rolling motion.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In dynamics, the moment of inertia plays a critical role in determining how much energy is associated with rotational motion, influencing how an object rolls on a surface.

Rolling Without Slipping Constraint

When an object rolls without slipping, there is a direct relationship between its translational speed and its angular speed, given by the condition v = ?r, where r is the radius. This constraint ensures that there is no relative motion between the point of contact and the surface, linking the two forms of kinetic energy.

Conservation of Mechanical Energy on an Inclined Plane

In the absence of non-conservative forces like friction doing work, the total mechanical energy (the sum of kinetic and potential energy) of a rolling object is conserved. When rolling uphill on an inclined plane, the object converts some of its kinetic energy into gravitational potential energy as it gains height. This energy transformation principle allows for the analysis of speed and energy changes along the incline.

Transcript

00:01 In this question, given data is that radius of hollow sphere is given, that is r is equal to 0 .15 meter.

00:16 And rotational inertia about a center, its center of hollow square is given, that is, value is i is equal to 0 .040 kg meter square.

00:31 And now in the question it is saying that this hollow sphere is, rolling down on an inclined surface having a slope that is with the horizontal that is 30 degree so consider that this is the roller square at this initial having position that is initial position is that is this and at this initial position the total kinetic energy is given that is k i is equal to 20 jule so this data is given and now you can see that in the question there are four parts given so i will ask all the parts one by one so let's start answer the given question so in the part a it is asking that we have to find the how much of its total kinetic energy is rotational so for finding this before i will try to find the mass of hollow spare so use this data that is given i is equal to 0 .040 kmeter square that means we know that the rotational inertia of hollow sphere about its center will be formula is that is 2 over 3 m r square is equal to i can say that it will be equal to 0 .040 where m is a mass of hollow sphere and r is a radius of hollow sphere so now next step is here put the value of r that is radius given 0 .15 meter in the left side and so simplify it then i will get the m that is mass of hollow square that is equal to after solving final value is that is 2 .7 kg so now next step is we know that the k rotational will be written as that is half i omega square and total kinetic energy will be written as that is at initial position that is we know that it is sum of translation kinetic energy and rotational kinetic energy.

03:08 So it means i can say that k total will be equal to half mv square plus half i omega square.

03:19 So now next step is k rotational divided by k total will be equal to that is half i omega square.

03:36 So i will be use this formula that is i is equal to for whole sphere.

03:43 About its center that is i can say that 2 over 3 m r squared multiplied by omega square divided by half m now v will be written as that is we know the relation that is we will be equal to linear speed will be equal to angular speed multiplied by radius so that means here i am putting the value of v that is equal to omega r then square plus half i will be written as again i can say that replace it by 2 over 3 m r square times of omega square now simplify it then i will get the ratio of total rotational kinetic energy to total kinetic energy is equal to 0 .40 so now in the question it is asking that we have to find the how much is of its total kinetic energy is rotational so i can see that it will written as that is k rotational is equal to 0 .40 multiplied by k total so initially the total kinetic energy is given that is 20 jule so that means i can say that here i will write that is 0 .40 multiplied by 20 jule so simplify it then i will get the answer that is for part a that is equal to 8 jule so this is the answer of part a that means k total rotational kinetic energy is 40 % of total kinetic energy and a value will be that is 8 jule now come to the next part that is part b in which it is asking that we have to find the speed of center of mass of whole sphere at the initial position so for finding this we know that you use the data from part a that is i can say that the total rotational kinetic energy at initial position will be equal to 8 jule and we know that the formula for k rotational is which i had written in the part 8 that is k rotational is equal to half i omega square so i am writing this term on the right left side that is half i omega square is equal to 8 jule now next step is i will substitute the data that is given i is given in the that is rotational inertia about its center of hollow square is given that is 0 .040 multiplied by omega squared will be written as by using this relation i can say that omega will be equal to v over r so that means v over r will be that is radius of holosphere that is given value that is 0 .15 meter whole square is equal to 8 now simplify it then i will get the answer for part b that is speed of center of mass of holders pair at initial position will be equal to after simplifying value will be that is 3 meter per second so this is the answer for part b now come to the next part that is part c in which it is saying that holosper is moving up and covering a 1 meter inclined distance that means in the figure i can say that this is the initial position of always spare so after it it is moving up and covering a one meter distance so consider that this is the one meter distance and it is reaching at this point so height will be written as that is i can say that this height will be equal to one times of sine 30 degree that means i can say that 0 .50 meter and in the question it is asking that for part c that is after reaching at this end point we have to find total kinetic energy so for finding this total kinetic energy at this end point i will use the law of conservation of energy so it will be written as it is initial total initial kinetic and sorry total initial energy is equal to total energy at final point so that means i can say that k initial will be written as here i'm taking a difference level that is passing through its initial position that means at this point at this line taking after taking this difference i can say that the total initial energy will be written as that is half sorry ki plus ui is equal to i can say that total final energy will be written as it is kf plus uf that means initial kinetic energy plus initially potential energy is equal to final kinetic energy plus potential energy at final point so after taking a reference at its initial point reference level which is passing through it initial point then i can say that this will become zero and k i is given in the cushion that is 20 jule is equal to kf plus uf will be written as that is m g h now put the value of m that is i had find in the part a already found find that is m is equal to 2 .7 kg and g that is 9 .80 meter per second square and h will be which i had written in the figure that is h is equal to 0 .50 meter so i can say that it will be written as that is 20 is equal to kf that means total kinetic energy at end point where it is after covering a one meter inclined distance with respect to initial position plus m g s m will be that is 2 .7 kg multiplied by g will be 9 .80 meter per second square multiplied by h will be that is 0 .50 meter so now simplify then i will get the answer for part c that means total kinetic energy will be equal to 6 .9 joule.

11:30 So that means i can say that after this always fail when moving up by 1 meter then in the and covering inclined distance that means then the total kinetic energy at end point will be written as that is kf will be equal to 6 .9 jules.

11:52 Now come to the last part that is part d.

11:55 So in the part d it is asking that we have to find the speed of center of mass at this end point at this point.

12:08 So for finding this, here i will use the line which is i had written in the part a that is i can say that k rotational, that means rotational kinetic energy is equal to 40 % of k total...

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