Question

points): Figure shows a wheel with inner radius a 10 cm and outer radius b-25 cm tha o through O Three forces produce a constant net torque on the wheel and rotate it starting from rest. ) Find the torque produced by each force on the whee b) If the wheel rotates with constant angular 1 (show your calculations explicitly). F- 100 N acceleration of magnitude a 3 rad/s?, what is the rotational inertia of the wheel? When the work done by the net torque on the wheel is 122.8 J, how many revolutions does it make? e) 0° F-15N

Answer 1

SOLUTION

Given:

Radius a= 10 cm = 0.1m

Radius b= 25 cm = 0.25m

Force F1 = 15 N , F2 = 5N , F3 = 100 N

angular accelalarion= 3 rad/ sec^2

Equation of Torque is given as :

T ( Torque) = r F sin$\Theta$ ( $\Theta$= angle b/w Force vector and extension of Radial vector at the application of force on periphery of wheels )

( r = radius of wheel)

Part a)

Finding Torque

T1 = rF sin$\Theta$

=0.25*15*sin90° ( r= a = .25m, $\Theta$=90°)

= 3.75 N- m ( clockwise)

T2= r F sin$\Theta$

= 0 ( $\Theta$= 0°)

T3 = rFsin$\Theta$ ( $\Theta$= 90°-40°= 50°)

= .1* 100 sin 50°

= 7.66N-m ( anticlockwise)

part (b)

Rotational inertia

​​​Tnet= I $\alpha$ ( I = moment of inertia about axis o, $\alpha$= angular accelalarion........eqn 1)

Now Tnet =( 7.66-3.75) N-m

= 3.91N-m ( anticlockwise)

From eqn 1

3.91= I * 3

I = 1.30 Kg- m^4

Part C)

Given = work = 122.8 J

Work done = Net Torque* $\Theta$

122.8= 3.91*$\Theta$

$\Theta$= 31.40 radian

Number of revolution= $\Theta$/ 2π

= 31.40/2π

= 4.9974 = 5 revolution