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poortech

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**1. A mass of 0.5 kg suspended from a flywheel, if the mass is released from rest and falls a distance of .5m in 1.5s calculate**

a) the linear acceleration of the mass

b) the angular acceleration of the wheel

c) the tension in the rope

d) the frictional torque, resisting motion

a) the linear acceleration of the mass

b) the angular acceleration of the wheel

c) the tension in the rope

d) the frictional torque, resisting motion

mass of wheel = 3kg

outside radius of wheel = 300mm

radius of gyration = 212mm

**2. Homework Equations**

I = Mk^2

K.E. = 1/2 Iw^2

θ = 1/2at^2

T = work done/angular desplacement

T= m(a+g)

I = Mk^2

K.E. = 1/2 Iw^2

θ = 1/2at^2

T = work done/angular desplacement

T= m(a+g)

**3. The Attempt at a Solution**

a)

v = .5/1.5 = .33m/s

a = v-u/t

.33/1.5 = .22m/s^2

b) angular acceleration = a/r

0.22/0.3 = .73m/s^2

c) T = m(a+g)

0.5(.22+9.8) = 5.01 N

d)Frictional Torque

moment of interia I = mk^2

0.3 x (.212)^2 = .135 kg.m^2

displacement θ = 1/2 at^2

1/2 (0.73)(1.5)^2 = .821 rad

K.E = 1/2 Iw^2

1/2 (.135) (1.1)^2 = .0816J

T = work done / angular displacement

.0816/0821 = .0993 Nm

a)

v = .5/1.5 = .33m/s

a = v-u/t

.33/1.5 = .22m/s^2

b) angular acceleration = a/r

0.22/0.3 = .73m/s^2

c) T = m(a+g)

0.5(.22+9.8) = 5.01 N

d)Frictional Torque

moment of interia I = mk^2

0.3 x (.212)^2 = .135 kg.m^2

displacement θ = 1/2 at^2

1/2 (0.73)(1.5)^2 = .821 rad

K.E = 1/2 Iw^2

1/2 (.135) (1.1)^2 = .0816J

T = work done / angular displacement

.0816/0821 = .0993 Nm

Plz let me know if I am doing it right, specially Frictional Torque...