Question 5 A hollow wheel of mass given by m = 50 ext kg...

Question 5 A hollow wheel of mass given by $m = 50 ext{ kg}$ rolls down an incline of $30^circ$ without slipping. The inner radius of the wheel is given by $R = 50 ext{ mm}$ and its thickness is given by $d = 5 ext{ mm}$. Gravity acts down. (a) Show that the moment of inertia of the wheel is given by $ar{I} = 138,125 ext{ kg mm}^2$. (6 marks) (b) Show that the linear acceleration of the wheel's centre of gravity $ar{a}$ is related to the angular acceleration of the wheel $alpha$ via $ar{a} = (55 ext{ mm})alpha$. (6 marks) (c) Analyse an appropriate free-body diagram to determine the linear acceleration of the wheel's centre of gravity $ar{a}$. (8 marks)

Transcript

00:01 Hi, in this question it is given that mass m is equal to 50 kg and r is equal to 50 mm and d is equal to 5 mm and in the first part of the question we need to find the moment of inertia of the hollow v.

00:25 So this i will be equal to half of m times r square plus r plus d whole square.

00:35 So this is equal to half of, so this m is 50 kg and then multiplied by r is again 50 square plus r plus d will be 55 square.

00:47 So this will be kg mm square.

00:51 So on solving this we will get 138125 kg mm square and this is moment of inertia and this is the answer for the first part of the question.

01:05 Now in the second part from the no slipping condition is cover the distance x equal to r theta.

01:21 So in this case this r will be r plus d theta.

01:28 Here we can write for our given case.

01:33 So this x will be equal to 55 theta and if you differentiate both side that is dx over dt this will be equal to 55 d theta d time.

01:46 Now if you differentiate this again so this is double derivative of x with respect to time equal to 55 multiplied by double derivative of theta with respect to time.

01:58 So this double derivative of position is known as acceleration and this will be equal to 55 times and the double derivative of angular displacement will be angular acceleration.

02:10 So this is the answer for the second part of the question.

02:16 Now in the third part here we have inclined plane which is at angle of 30 degree and here this is the wheel and here the weight force will be acting.

02:32 In this direction we will have mg cos 30 degree and in this direction we will have mg sin 30 degree and in this direction there will be coefficient of friction and it will be moving in this direction with acceleration a and on this mass there will be acting a normal force...

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