A small sphere of mass m and radius r is released from rest...

A small sphere of mass $m$ and radius $r$ is released from rest at $A$ and rolls without sliding on the curved surface to point $B$ where it leaves the surface with a horizontal velocity. Knowing that $a=1.5 \mathrm{m}$ and $b=1.2 \mathrm{m}$, determine $(a)$ the speed of the sphere as it strikes the ground at $C,(b)$ the corresponding distance $c .$

Key Concepts

Projectile Motion

Once the sphere leaves the curved surface with a horizontal velocity, it behaves like a projectile. The study of projectile motion separates the horizontal and vertical components of motion, where the horizontal velocity remains constant (neglecting air resistance) and the vertical motion is influenced by gravitational acceleration. This concept is fundamental for determining the time of flight and horizontal displacement before the sphere strikes the ground.

Rotational Dynamics

Rotational dynamics involves the analysis of the rotational motion of the sphere, particularly the relationship between torque, moment of inertia, and angular acceleration. The moment of inertia for a sphere (typically I = (2/5)mr²) plays a key role in partitioning the energy between rotation and translation, affecting the sphere's final speed.

Conservation of Mechanical Energy

In systems where non-dissipative forces are present—like friction acting statically in rolling without slipping—the mechanical energy is conserved. The gravitational potential energy lost as the sphere rolls down the curved surface is converted into both translational and rotational kinetic energy. This principle allows for the calculation of the sphere’s speed at different points of its motion.

Rolling Without Slipping

This concept ensures that the sphere’s translational and rotational motions are linked through the relation v = ?r, meaning the point of contact is momentarily at rest relative to the surface. Understanding this is crucial because it guarantees that energy is split between translational kinetic energy and rotational kinetic energy without any loss due to sliding friction.

Transcript

00:01 Hello friend at show in the figure a small sphere of mass am and radius r is released from rest from position a it rolls down without sliding that is sphere is performing pure rolling motion leave the surface with horizontal velocity if value of a is given 1 .5 meter and value of v is given 1 point this is not h i have marked wrongly this is b which is 1 .2 meter calculate part 8 velocity at c striking the ground and corresponding distance c this is b point this is c point for motion from a to b work done by gravity is m d g into a.

02:17 At a, kinetic energy is 0 and at b it is half m v square plus half i omega square since it is a spherical body.

02:40 So moment of inertia of spherical body we have to use about its axis will be 2 by 5 m r square and for pure rolling v skull to r omega.

03:01 So that these two concepts we have to apply.

03:06 So calculating the kinetic energy at b, half mv square, half i, that is 2 by 5 mr square into v square by r square.

03:33 So it would be 7 by 10 mv square.

03:41 Applying work energy theorem, work done in the position a to b is called to change in kinetic energy.

04:05 So work done from position a to b is given mg into a change in kinetic energy.

04:22 T2b have calculated 7 by 10 mb square minus 0.

04:29 So velocity at v, position b will get 4 .58 -49 meter per second.

04:47 For motion b2c, it is a projectile motion at 3 .5 motion at 3 .5 .4 .9 meter per second.

05:00 For motion, it is a projectile motion...

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