Solutions manual for mechanics and thermodynamics

1.In a game of billiards, the player wishes to hit a stationary target ball with the moving projectile ball. After the collision, show that the sum of the scattering angles is 90o. Ignore friction and rolling motion and assume the collision is elastic. Also both balls have the same mass.

SOLUTION The collision occurs as shown in the Øgure. We have m1 = m2 ¥ m.

y

vP

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Momentum conservation is:

p~P i = p~P + p~T

and we break this down into the x and y directions. Momentum conservation in the y direction is:

Momentum conservation in the x direction is:

We now have 3 simultaneous equations which can be solved. This involves a fair amount of algebra. We can do the problem much quicker by using the square of the momentum conservation equation. Use the

but the masses cancel out, giving

vP2 i = vP2 + vT2 + 2vP vT cos(µ + Æ)

which, from energy conservation, also equals

vP2 i = vP2 + vT2

implying that

cos(µ + Æ) = 0

which means that

µ + Æ = 90o

1.Show that the ratio of the angular speeds of a pair of coupled gear wheels is in the inverse ratio of their respective radii. [WS 13-9]

SOLUTION

2.Consider the point of contact of the two coupled gear wheels. At that point the tangential velocity of a point on each (touching) wheel must be the same.

v1 = v2

) r1!1 = r2!2

)!1 = r2 !2 r1

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3. Show that the magnitude of the total linear acceleration of a point

moving in a circle of radius r with angular velocity ! and angular p

acceleration Æ is given by a = r !4 + Æ2 [WS 13-8]

SOLUTION

The total linear acceleration is given by a vector sum of the radial and tangential accelerations

q

a = a2t + a2r

where the radial (centripetal) aceleration is

ar = v2 = !2r r

and

at = rÆ

so that p p

a = r2Æ2 + !4r2 = r !4 + Æ2

4.The turntable of a record player rotates initially at a rate of 33 revolutions per minute and takes 20 seconds to come to rest. How many rotations does the turntable make before coming to rest, assuming constant angular deceleration ?

SOLUTION

=3:46 rad sec°1

!= 0

t = 20 sec

number of rotations = 34:6 radian = 5:5 2ºradian

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5.A cylindrical shell of mass M and radius R rolls down an incline of height H. With what speed does the cylinder reach the bottom of the incline ? How does this answer compare to just dropping an object from a height H ?

SOLUTION

Conservation of energy is

Ki + Ui = Kf + Uf

0 + mgH = 12mv2 + 12I!2 + 0

For a cylindrical shell I = mR2. Thus

mgH = 12mv2 + 12mR2!2

and v = r! giving (with m cancelling out)

gH = 12v2 + 12R2(Rv )2

=12v2 + 12v2

=v2 p

) v = gH

p

If we just drop an object thenpmgH = 12 mv2 and v = 2gH. Thus the dropped object has a speed 2 times greater than the rolling object. This is because some of the potential energy has been converted into rolling kinetic energy.

6.Four point masses are fastened to the corners of a frame of negligible mass lying in the xy plane. Two of the masses lie along the x axis at positions x = +a and x = °a and are both of the same mass M. The other two masses lie along the y axis at positions y = +b and y = °b and are both of the same mass m.

A)If the rotation of the system occurs about the y axis with an angular velocity !, Ønd the moment of inertia about the y axis and the rotational kinetic energy about this axis.

B)Now suppose the system rotates in the xy plane about an axis through the origin (the z axis) with angular velocity !. Calculate the moment of inertia about the z axis and the rotational kinetic energy about this axis. [Serway, 3rd ed., pg. 151]

SOLUTION

A) The masses are distributed as shown in the Øgure. The rotational inertia about the y axis is

X

Iy = ri2mi = a2M + (°a)2M = 2Ma2

i

(The m masses don't contribute because their distance from the y axis is 0.) The kinetic energy about the y axis is

. m