c a hollow sphere mass m radius r rolls down a 45 degree incline let be the angle of rotation of

Step 1: The total kinetic energy of the sphere can be written as the sum of its translational ki

C a hollow sphere mass m radius r rolls down a 45 degree...

(C) A hollow sphere (mass = m, radius = r) rolls down a 45-degree incline. Let φ be the angle of rotation of the sphere, and ψ = the angular velocity (θ)/(t) = αr (φ)/(t)αr (δ). (i) Write the total kinetic energy of the sphere in terms of ψ. (ii) Write the gravitational potential energy in terms of φ. (iii) The total energy has the form (1)/(2)μψ^(2)-Γφ, so use part A to determine the angular acceleration. (expression for K = (5)/(6)m*r^(2)*ψ^(2) You are correct. Your receipt no. is 158-3210.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 So here we are given that a hollow sphere of mass m and radius r rolls down a 45 degree incline.

00:10 If phi is the angle of rotation, right? so here you have got phi angle of rotation and psi is the angular speed, right? angular velocity, let's say, right? now you have to write down the total kinetic energy of the sphere in terms of psi.

00:31 So it's pretty simple.

00:33 The total kinetic energy will be equal to kinetic energy which is a component of the rotation plus kinetic energy which is a translational component right rotational component is given as high half i angular sphenosity which is psi square plus translational kinetic energy you all know it's half mv square right visa velocity and this is half times moment of inertia i is for a hollow sphere is 2 3rd m r square psi square plus half m again v is psi r psi right whole square of that and if you evaluate this will be 5 by 6 m r square psi square right is the total kinetic energy next you have to calculate the gravitational potential energy in terms of psi so the the potential energy is equal to mgh and this height h this is 45 degree side in terms of d will be mgh is d sine 45 degree and mg times this d can be written as r phi right r times the radius times the angle of rotation gives you the total distance it covered times sine 45 degrees next you have to find the expression of the angular acceleration right now if you have a sphere which is rolling down right so here you need to calculate the component of the acceleration due to gravity which is g sine theta right and theta is 45 degrees so the acceleration will be g sine 45 degrees e will be equal to r alpha which is g sine 45 degrees right and alpha from here will be equal to g by r sine 45 degrees is the angular acceleration expression right sorry i think you can do the other way around just let me help you then you know you can tackle it from the torque equation because since it's a case of rolling right pure rolling so here you can apply the formula the torque tau will be equal to i alpha right.

03:05 Now if you see the torque here which is nothing but if you see the component of this weight about here you have the so here you have got mg sine theta acting right.

03:17 The weight is acting downwards and the torque will be mg sine of 45 degrees times r right.

03:26 Right.

03:26 R is the radius at the point of contact...

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