7) Suppose a wheel of moment of inertia about its centre of 2 kgm^2 is spinning

Question

7) Suppose a wheel of moment of inertia about its centre of 2 kgm^2 is spinning with angular velocity of 15 rads^-1. If it is brought to rest by a steady braking torque T in 5 sec, what is the value of T? 8) State the principle of conservation of linear momentum? ii)An object of mass 2kg is moving with a velocity of 3ms1 and collides head on with an object B of mass 1 kg moving in the opposite direction with a velocity of 4ms^-1. After collision both objects stick so that they move with a common velocity V. Calculate V.

Explanation / Answer

7. use the formula for Torque T = I A

where I is moment of inertia = 2 kgm^2

A is angular accleration = W/t = 15/5 = 3 rad/s^2

so

Torque T = 2 * 3 = 6 Nm
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According to the law of conservation of momentum , total momentum of a a system of masses is constant

i.e total intial momentum = total final momentum

ii) use the formula

m1u1 + m2u2 = m1V + m2V

where m1 = 2kg

u1 = 3 m/s

m2 = 1 kg

u2 = 4 m/s

V is the common velocity after sticking together

so

V(m1+m2) = 2 * 3 - 1*4

V(2+1) = 2

V = 2/3 = 0.67 m/s is the common velocity

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