Determine the kinetic energy of the space probe of Prob....

Key Concepts

collision_dynamics

Collision dynamics examines the conservation laws governing interactions between bodies during impacts. Key conservation principles, such as linear momentum and angular momentum, are applied to determine the subsequent motion of objects, including any induced rotation after a collision.

rotational_motion

Rotational motion occurs when an object spins about an axis, typically its center of mass. The rotational kinetic energy is determined by the moment of inertia and the angular velocity of the object, making it an essential consideration when analyzing post-collision dynamics that involve spinning or turning about a mass center.

center_of_mass

The center of mass is the point in a body or system where the mass is considered to be concentrated. When analyzing the motion after a collision, it is useful because the movement of the system can be divided into translation of the center of mass and rotation about the center of mass, simplifying the analysis.

kinetic_energy

Kinetic energy is the energy an object possesses due to its motion. In analyzing moving objects, both translational kinetic energy, given by (1/2)mv², and rotational kinetic energy, given by (1/2)I?², are considered. This concept is fundamental for calculating the energy of objects in either linear or rotational motion.

Transcript

00:01 Hi friends as soon as the diagram there is space probe having the weight of 3 ,000 pounds and radius of variation kx k y k z and its value are given.

00:12 Its initial angular velocity is zero.

00:15 If a meteorite having the mass of 5 oens striking to the panel of the probe at point a and penetrated to the other side without changing the direction, only magnitude gets reduced.

00:31 By 25%.

00:33 If vital angular velocity of probe is this, we have to calculate kinetic energy of the probe after impact.

00:47 It is also given that change in velocity of mass center of the probe along x direction is minus 6 .75 inch per second.

00:58 Let us start doing it.

01:05 Mass of the space probe and weight divided by acceleration due to gravity so it will be 93 .17 slugs mass of meteorite m 5 o's upon 16 into 32 .2 so it would be 0 0909705 slugs position vector of point of impact a or a will be 9 i cap plus 0 .75 k cap fit initial linear momentum meteor right mv not is called to 0 .009705 into vx i cap plus v y j cap plus v z k cap now any moment of momentum about origin initial initial angular momentum with you write about origin will be r a cross m b not.

04:27 M is constant so you can take outside of determinant while solving the cross product i cap j cap k cap 9 .75 so on solving it, bx, b5, vz.

04:48 So on solving it, this value you will get 0 .009705 into minus 0 .75 b by i cap plus 0 .75 bx minus 9 v z j cap plus 9 v y k cap say this is equation number two final linear momentum you can find since 75 % it lost so it will be 0 .75 mb0 so when we takes 0 .7 only this quantity will change so it becomes 0 .00729 px i cap plus vy j cap plus v .z k cap.

06:27 Final angular momentum of meteorite about origin.

07:04 This is not and this is final.

07:10 So r a...

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