1. A wheel with a rotational inertia of 5.0 kg m 2 is rotating in a horizontal p

Question

1. A wheel with a rotational inertia of 5.0 kg m2 is rotating in a horizontal plane at 600 rev/min. A rim brake exerts a torque of 500 Nm on the inside rim of the wheel. Find the initial angular velocity in units of rad/s and the number of revolutions the wheel rotates through before it comes to rest.

2. A uniform metal ball of mass 31 g and radius 1.4 cm rolls down (without slipping) a slope of height 32 cm from rest. Find its linear (c.o.m.) speed and angular speed at the bottom of the slope.(The rotational inertia of a uniform ball is I = 2/5 M R2 .)

3. A satellite of mass 1200 kg moves uniformly at the height of 380 km in a circular orbit around the Earth. Find

a) the gravitational force on the satellite;

b) the speed of the satellite;

c) the period for the satellite around the Earth.

(The mass of he Earth: Mearth = 5.98 x 1024 kg; the radius of the Earth: Rearth = 6370 km . )

4. The volume rate of a flow in a horizontal pipe is 4.0 L/s. The diameters of the pipe at points A and B are 2.0 cm and 5.0 cm, respectively. Find

a) the speeds of the flow at points A and B;

b) the difference in pressure between the two points, i.e. Pb - Pa

Explanation / Answer

1) 1 rev = 2*pi radians

1min = 60 s

a) w = 600rev/min = (600*2*3.14)/60 = 62.8 rad/s

Torque = I*alfa

alfa = 500/5 = 100 rad/s^2

angular displacement = (wf^2 - wi^2)/2*alfa

angular displacement = (62.8*62.8)/(2*100) = 19.7192 radians

angular displacement = 19.7192 / 6.28 = 3.14 revolution

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2) E1 = PE = m*g*h

KE = KEr + KEt = 0.5*I*w^2 + 0.5*m*v^2

KE = 0.5*(2/5)*m*R^2*v^2/R^2 + 0.5*m*v^2

KE = 0.5*(7/5)*m*v^2

KE = PE

0.5*(7/5)*m*v^2 = m*g*h

v = sqrt((10/7)*g*h)

v = 2.12 m/s

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3)

a) F = GMm/r^2

F = (6.67e-11*5.98e24*1200)/(6370000+380000)^2

F = 10505.11 N

b) v = sqrt(GM/r) = sqrt(6.67e-11*5.98e24)/(6370000+380000)

v = 7687.08 m/s = 7.68 km /s

c)

T = (2*pi*r)/V = (2*3.14*(6370000+380000))/7687.08

T = 5514.45 s = 1.532 hours

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4)

continuity equation

a) at point A

area*va = 4

pi*1e-4*va = 4e-3

va = 12.74 m/s

at point B

area*vb = 4

pi*2.5*2.5*1e-4 = 4e-3

vb = 2.04 m/s

c)

bernoullis pricilple

Pa + 0.5*rho*va^2 = Pb + 0.5*rho*vb^2

Pb - Pa = 0.5*rho*(va^2 - vb^2)

Pb -Pa = 0.5*rho*((12.74*12.74)-(2.04*2.04)) = 79.073*rho pascals

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