P04-04 The speed of a rotating system is controlled with a brake as shown in Fig

Question

P04-04 The speed of a rotating system is controlled with a brake as shown in Fig. and a radius of gyration with respect to the axis of rotation of 200 mm. The kinetic coefficient of friction between the brake pad and brake drum is 0.50. When a force of 500 N is being applied to the brake lever, determine the horizontal and vertical components of the reaction at support B of the brake lever and the time t required to reduce the speed of the system from 1000 rpm to The rotating parts of the system have a mass of 300 kg 160 mm 100 mm Brake lever Brake pad Rotating system 260 mm Brake drum l rest.

Explanation / Answer

Solution:

m= 300 kg ; K= 200mm ; Radius of Drum is 260 mm ; No=1000 rpm ; Nf=0

wo = 2 x pi x 1000/60

= 104.71

coff of friction is 0.5

Moment of inertia is given by I = mk^2

= 300 X (0.2)^2

= 12

Torque is given by T = I w

= 12 X 2 pi (1000)/60

= 1256.63 N-m

tangential force is given by T/R

Ft = 1256.63/(0.26)

= 4833.21 N

Normal reaction at centre of drum is given by N = Ft/(coff of friction)

= 4833.21/0.5

=9666.4 N

At rest, system will be at equillibrium

Sum of horizontal & vertical forces will be zero.

H =0

Ft + Horizontal component at B

Horizontal component at B = - Ft

= - 4833.21 N (heading towards right)

Similarly

V=0

N -500 + Vertical component at B = 0

Vertical component at B = 500 -N

= 500- 9666.4

= -9166.4 N (downward direction)

Torque applied at drum is 1256.33 Nm

therefore T = F (average velocity) x t

1256.33 = 4833.21x (104.71 + 0)/2 x R x t

t = 0.01 s

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