A simple band brake is applied to a shaft carrying a flywheel, of mass 250kg and radius of gyration 300mm. The shaft speed is 200rpm. The drum diameter is 200 mm and the coefficient of friction 0.25. The free end of band is attached at 100mm from the fulcrum and effort of 120 N is applied on lever at 280 mm on the other side of the fulcrum. The angle embraced by belt is 225 degrees. Determine for counter clockwise rotation of -
(i) Braking torque - (ii) The number of turns of flywheel before it comes to rest. -

Mumbai University > MECH > Sem 5 > Theory Of Machines 2

Marks: 7 M

Year: Dec 2014

Given:

M=250kg

K= 300mm = 0.3m

N = 200rpm

d = 200mm; r = 100mm = 0.1m

μ=0.25

a = 100mm = 0.1m

F = 120N;

L = 280mm = 0.28m

$Θ=225^0 = 225 \times π /180 = 3.93 rads$

Formula:

$T1 = T2 e^{μθ}$

T = (T1-T2) r

Solution:

$Ta = Tb e^{μθ} \\ Ta = Tb e^{0.25 \times 3.93} = 2.67 \times T2 --------i$

Taking moment about O

Ta x A = F x L

Ta x 0.1 = 120 x 0.28

Ta = 336 N

Tb = 125.8 N

T = (Ta-Tb) x r = (336-125.8) x 0.1 = 21Nm

Number of turns:

T $= Iα \\ = Mk^2 α$

$21 = 250 \times 0.3^2 x α \\ . α = 0.933 rad/s^2$

$W1 = 2πN/60 = 2 \times 3.14 \times 200 /60 = 20.94 rad/s \\ W_2^2 = w_1^2 + 2αφ \\ 0 = 20.94^2 - 2 \times 0.933 \times φ . φ = 235 rad$

No. of turns $= φ/2π = 37.42 \ \ turns$