A 2 -kg solid sphere of radius r=40 mm is dropped from a height h=200 mm and lands on a uniform

A 2 -kg solid sphere of radius r=40 mm is dropped from a...

A 2 -kg solid sphere of radius $r=40 \mathrm{mm}$ is dropped from a height $h=200 \mathrm{mm}$ and lands on a uniform slender plank $A B$ of mass $4 \mathrm{kg}$ and length $L=500 \mathrm{mm}$ that is held by two inextensible cords. Knowing that the impact is perfectly plastic and that the sphere remains attached to the plank at a distance $a=40 \mathrm{mm}$ from the left end, determine the velocity of the sphere immediately after impact. Neglect the thickness of the plank.

Key Concepts

Rigid Body Dynamics

Rigid body dynamics deals with the motion of objects that do not deform during movement, particularly focusing on both translational and rotational motions. In collision problems involving bodies like a plank, analyzing the coupled translational and rotational motion, including the impact of support constraints, is essential to determine post-collision behavior.

Moment of Inertia

The moment of inertia measures how a body's mass is distributed relative to an axis of rotation and indicates its resistance to angular acceleration. In problems involving collisions that lead to rotation, computing the moment of inertia for each component of the system (or the composite body) is key to applying the conservation of angular momentum.

Gravitational Energy Conversion

This concept involves the transformation of gravitational potential energy into kinetic energy as an object falls. In free-fall problems, knowing the height allows one to calculate the velocity just before impact by applying energy conservation, assuming negligible air resistance.

Perfectly Plastic (Inelastic) Collision

In a perfectly plastic collision, the colliding bodies stick together after impact. Although kinetic energy is not conserved due to energy loss (transformed into heat and deformation), momentum—more precisely linear or angular momentum along the appropriate directions—remains conserved.

Conservation of Angular Momentum

When external torques are absent or their effects cancel when taken about an appropriate pivot, the total angular momentum of a system remains constant during a collision. This principle is especially important in analyzing rotational effects when an object collides with a body capable of rotation or pivoted motion.

Transcript

00:01 In this question, the moment of inertia of a sphere i s equal 2 over 5 mass of the sphere times radius square so it will be equal 1 .28 times 10 power negative 3 and the units kilogram dot meter square and for the plank ap the moment of inertia equal 1 over 12 mass ab times lens square which will be equal 8 point or 83 point 33 times 10 power negative 3 kilogram dot meter square so velocity of the sphere at impact vs equals root 2 gh which which is equal 1 .98 meter per second.

01:18 So before impact, linear momentum mass of the sphere times velocity of the sphere equals 4 kilograms multiply 1 .98 equals 7 .92 kilogram kilogram.

01:40 Gram dot meter per second.

01:44 With its line of action lying at distance l over 2 minus a from the midpoint of the plank so l over 2 minus a equals 0 .25 minus 0 .04 which is equal 0 .21 meters and after impact the edge equal equals l, cosine 30 plus radius r, which is equal 0 .47 meters and hs is l over 2 minus a, which is equal 0 .21 meters.

02:36 So, c .h or cs equals root ch square plus h s square plus h square will be equal point 51 meters and then beta equals h s over ch will be equal point 44 so beta in this case equals 23 .94 degrees and vs equal c s times omega equals point 51 omega and vg equals l cosine 30 times omega equals 0 .43 omega.

03:41 By applying moment about point c we have mass of the sphere times vs multiply l over 2 minus a plus 0 equal ms v s multiply c s plus mass ab multiply v g multiply c h plus i s plus i ab which is equal also i f f times omega so the left hand side omega so the left hand side which is msvs l over 2 minus a will be equal 0 .83 kilogram dot meter square per second and for ic it will be equals 1 .37 kilogram dot meter square so substituting again in our equation so have 0 .83 equals 1 .37 omega so omega will be equal 0 .6 -07 radian per second and counterclockwise...

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