A solid sphere rolls down two different inclined planes of the same heights but different angles

A solid sphere rolls down two different inclined planes of...

Key Concepts

Acceleration on an Inclined Plane

The acceleration of a rolling object down an incline depends on the component of gravitational force along the incline, which is proportional to the sine of the incline angle, and also on the object's moment of inertia. A steeper incline increases the acceleration (due to a larger sine value), even though the final speed at the bottom remains the same if starting from the same height. This difference in acceleration means that the time taken to reach the bottom will vary with the incline angle.

Rotational Dynamics and Moment of Inertia

The moment of inertia is a measure of an object’s resistance to changes in its rotational motion. In rolling motion, the distribution of mass (captured by the moment of inertia) affects how much of the gravitational potential energy is converted into rotational kinetic energy versus translational kinetic energy. This interplay influences the acceleration down the incline but does not alter the final speed if no energy is lost.

Conservation of Energy

This concept explains that the gravitational potential energy of an object at a height is converted into kinetic energy as it moves down an incline. In rolling motion, this kinetic energy is divided into translational and rotational components. Since the initial potential energy is solely determined by the height, objects starting from the same height will have the same total kinetic energy at the bottom, regardless of the slope's angle.

Rolling without Slipping

Rolling without slipping means that the point of contact between the rolling object and the surface is momentarily at rest relative to the surface. This condition links the translational speed and rotational speed of the object. It ensures that there is no energy loss due to sliding friction, and the energy conversion from potential to kinetic energy occurs smoothly between translational and rotational forms.

Transcript

00:04 In this problem, a solid sphere is rolled from two different inclines having different angle of inclination but from same height.

00:15 So we have to comment about the speed with which it reaches the ground.

00:20 Will it reach with the same speed or with different speed? so before starting this problem, first of all, we should examine that from where velocity of this sphere, will come from.

00:36 So actually whatever is the potential energy of this sphere, this potential energy will be converted into kinetic energy of the sphere at the bottom.

00:45 So at the expense of potential energy, kinetic energy is increasing.

00:50 Now what is the potential energy at the top? so in both the cases, potential energy will be equals to m into g into h where m is the mass of the sphere.

01:05 Now, who is the what is the kinetic energy with which it reaches the ground? so kinetic energy will be, let us say, it reaches the ground with speed v1 and in this case it reaches with speed v2.

01:18 So there will be two types of kinetic energy.

01:21 One due to translational motion and the second one due to rotational because it is rolling.

01:26 So there will be angular speed associated with the spheres omega 1 and omega 2.

01:32 So in first case, total kinetic energy will be kinetic energy due to translation plus kinetic energy due to rotation.

01:45 So similarly the same formula will be applicable for both of these cases.

01:52 So i am writing both the case in parallelly.

01:56 Now what is the translational kinetic energy? so translational kinetic energy is how? m v1 square and rotational kinetic energy will be half i omega 1 square similarly here again we can write half m v2 square plus half i omega 2 square now what is the moment of inertia of a solid sphere about its diameter of about its diameter so for solid sphere moment of energy is 2 by 5 mr square so now we can write here this will be equal to half m v1 square plus half into 2 by 5 m r square into omega 1 square so we can solve this so omega 1 into r omega 1 is equal to v1 therefore we can write omega 1 square into r square and we can replace this term with b1 square so this will be half m v1 square plus 1 by 5 m v1 square so this is equal to this will be equals to lcm will be 10 and in numerator it will be 5 m v1 square plus 2m v1 square which is equals to 7m v1 square divided by 10 now similarly if we do for the second case so here only notation will change so if we solve this with the same procedure so it will so its result will be 7 mv2 square divided by 10 all right so these are the total kinetic energy in both the cases now we know that these kinetic energy are equal to potential energy so if we equate or if we apply energy conservation in first case then what we can write is 7m v1 square upon 10 should be equals to m g h so m can be cancelled.

04:05 So v1 will be equals to 10gh divided by 7 and square root over this term.

04:13 So this is the speed with which the sphere will reach at the ground.

04:21 Now we have to do the calculation for the second case.

04:25 So in second case we know that potential energy is same because height is same.

04:29 So this quantity, that is this total kinetic energy, half mv2 square upon 10, should be equal to the total potential energy...

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