Question

Time remaining 1.19.43 The 8-in.-radius brake drum is attached to a larger flywheel that is not shown. The total mass moment of inertia of the drum and the flywheel is 14 lb-ft-s2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.35. Knowing that the angular velocity of the flywheel is 360 rpm clockwise when a force P of magnitude 95 lb is applied to the pedal C, determine the number of revolutions executed by the flywheel before it comes to rest. (Round the final answer to one decimal place.) 6 in. А 10 in. Sin 15 in The number of revolutions executed by the flywheel before it comes to rest is

Answer 1

the drum a mes to rest. (Round the finala 6 in. flywheel is 360 rpm clockwise when a force P of magnitude 95 lb is a the flywheel before it comes to rest. (Round the final answer to one 6 in. А 10 in. N Dy 10 in. fix Sin. 10.35N 15 in. 0.35N The number of revolutions executed by the flywheel be 15 in.

Take moment of forces about pin-A;

$\sum M_A = 0$

→95(15 – 6) - N(10) – (0.35N)(8 – 6) = 0

$\Rightarrow N = 79.907\;\;lb$

For drum;

Σ MD MD = Iα

$\Rightarrow (0.35N)\left ( \frac{8}{12} \right ) = I\alpha$

$\Rightarrow (0.35\times79.907)\left ( \frac{8}{12} \right ) = 14\alpha$

$\Rightarrow \alpha = 1.332\;\;rad/s^2$

Angular velocity;

$\omega = \frac{2\pi N}{60} = \frac{2\pi (360)}{60} = 37.699\;\;rad/s$

Angle moved;

$\theta = \frac{\omega^2}{2\alpha} = \frac{37.699^2}{2(1.332)}$

$\Rightarrow \theta =533.582\;\;rad$

Rotations,

$N = \frac{\theta}{2\pi} = \frac{533.582}{2\pi}$

$\Rightarrow N =84.9\;\;revolutions$

...(Answer)