= 2t^3-12t^2, t>0. A wheel of radius 50cm rotates about a fixed axis. The angula

Question

= 2t^3-12t^2, t>0. A wheel of radius 50cm rotates about a fixed axis. The angular position of the wheel is given by the expression above. The wheel is a rigid body, which means that any point on the wheel traces out a circle as the wheel rotates. The moment of inertia is 10kg*m^2. I) What is the torque in nm on the wheel at 2 seconds? II) If the torque acting on the wheel is provided by a single force applied tangentially to the circumference of the wheel, what is the value(magnitude) in Newtons of this force at 5 seconds?

Explanation / Answer

the angular speed is

(t) = d / dt

(t) = d [ 2t3-12t2 ]/ dt = 6t2 - 24t

then,

a) the torque is

= I (2s)

= ( 10 kg.m2 ) [ 6( 2 s )2 - 24( 2 s ) ] = -240 N.m

b) the torque is

= I (5s)

= ( 10 kg.m2 ) [ 6( 5 s )2 - 24( 5 s ) ] = 300 N.m

but, torque can be defirned like

= r F sin

if the force is applied tangentially to the circunferen then, = 90° so

= r F

hence

F = / r = ( 300 N.m ) / ( 0.5m ) = 600 N

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