1. (10 points) The angular velocity of a rotating disk is...

1. (10 points) The angular velocity of a rotating disk is given by ? (t) = 1.6t³ - 4.2t² (rad/s). What is the average angular acceleration of the disk between t = 1 s and t = 3 s? Answer: 4 rad/s² 2. (10 points) The angular position of a point on a rotating wheel is given by ?(t) = t? - 4t² + 6, where ? is in radians and t is in seconds. What is the point's angular acceleration at t = 2 s? Answer: 40 rad/s² 3. (10 points) Time dependent angular velocity of a particle is given by ?(t) = 5t² + 6t + 1 (rad/s). What is the average angular acceleration in the time interval t=0 to t=2s? Answer: ???? = 16 rad/s² 4. (10 points) Two point particles are fastened to a rod of negligible mass as shown in the figure. If the rotational inertia about an axis that is perpendicular to the rod and 20 cm away from the first particle is I = 0.09 kg.m², what is the distance x? from second particle to the rotation axis? m? = 4 kg x? = ? x? = 20 cm m? = 2 kg Answer: x? = 0.05m 5. (10 points) A rotor has a moment of inertia of 4.20×10?² kg.m². How much energy is required to bring it from rest to 9300 rpm (rev/min) speed?

Transcript

00:01 Hello, the question is taken from physics and the question is the angular velocity of a rotating disk is given by.

00:06 So given that omega is equal to 1 .6 t cube minus 4 .2 t squared.

00:13 So what will be the value of the average angular acceleration of the disk between t is equal to 1 to 3 second? so we know that alpha is equal to omega to minus omega 1 divided by t 2 minus t 1.

00:26 So that will be 1 .6 into 3 cube minus 4 .2 into 9 minus 1 .6 plus 4 .2 divided by 2.

00:40 So from here the value of alpha will be 1 .6 into 9 minus 4 .2 into 9 sorry not 20.

00:50 So if you solve it you get the value of 4 radiant per second square and the second part is so let me value at the second part.

00:59 The angular position of a rotating wheel is given by theta t, where theta is the radian in t second, what is the point's angular acceleration at t is equal to two second? so theta is equal to t4 minus 40 square plus 6.

01:17 So from here, the value of angular acceleration will be d2 theta over d2 square.

01:24 So let us first evaluate d theta over dt, which is 40 cube minus.

01:30 So further differentiating it we get 12 t square minus 8 at t is equal to 2 the value of alpha will be 12 into 4 minus 8 48 48 minus 8 is 40 radian per second square which is the required value of the angular acceleration time dependent angular velocity of the particle is given as omega t is equal to 5 t square plus 60 plus 1 that is the required value of the angular velocity what is the average angular acceleration in that time interval 0 to 2 second okay so angular acceleration is omega t2 minus omega t 1 divided by t 2 minus t 1 so that will be at t 2 0 to 2 second so 5 into 4 20 plus 12 plus 1 minus at 0 will be 1 so divided by t 2 minus t 1 that is 2 minus 0 that is 2 32 by 2 is 16 radian per second square which is the required value of the angular acceleration next is to point particles are fastened to the road of negligible mass as shown in the figure if the rotational inertia about the axis that is perpendicular to the road and 20 centimeter away from the first particle is i is equal to 0 .09 kilogram per meter square what is the distance x2 from the second particle of the rotation axis so this can be solved by moment balance which says that m2 x2 is equal to m1 x1 okay the value of m2 is 4 into x2 that is m1 x1 is 20 into 2 so from here x2 is equal to 10 centimeters one second whether i did anything or m to x1 whether it is m2 into x1 80 is equal to 2 x2 no 4 into 0 0 0 0 0 0 4 into 20 centimeter away from the first particle is i omega square what is the distance x2 from the rotation axis so let me evaluate this value let me each now given the value of the total moment of inertia, so we know that i is equal to half m1r1 square, which is x1, plus half m2 x2 square, that is 0 .09, which is taking half common...

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