A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 πNm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is…

Question

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 πNm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is _________m.

Accepted Answer

Correct answer: 40. (40) Mass of the disk, m=1 kg. Initial angular velocity, _i=1800 rpm. Final angular velocity, _f=2100 rpm. External torque, _ ext=25 πNm. Time, t=40 seconds. First, convert the rotational speeds from revolutions per minute (rpm) to radians per second (rad/s): _i=1800 2 60=60 πrad/s _f=2100 2 60=70 πrad/s Next, use the equation of motion for rotation to find the angular acceleration : _f= _i+ αt70 =60 + (40 ) Solving for : = 70 -60 40= 10 40= 4 rad/s^2 The torque and moment of inertia relationship is given by: =I For a thin solid disk rotating along its diameter, the moment of inertia I is mR^24. Thus: 25 = 1 R^24 4 Solving for R : 25 = R^216R^2= 25 16 =400R=20 m The diameter of the disk is: Diameter =2R=2 20=40 m

$A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of 25 πNm for 40 s, the speed increases to 2100 rpm. The diameter of the disk is _________m.$

Solution

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