# HW2 for Chap11 ```PHY126 HW Set 2 (Chap.11)
Problem 11.42
Description:A physics student is standing on an initially motionless, frictionless turntable with
. She's holding a wheel of rotational inertia 0.22
spinning at 130
rotational inertia 0.31
about a vertical axis, as in the figure. When she turns the wheel upside down, student and turntable
.
begin rotating at 70
Part A
What is the student's mass, considering her to be a cylinder 30
in diameter?
Part B
How much work did she do in turning the wheel upside down? Neglect the distance between the axes
of the turntable and wheel.
Solutions
(a) Since the turn table is frictionless, there are no external torques about its axis, and the z component of
angular momentum is conserved. Initially this is all spin of the wheel, L0 = (0.22 kg ⋅ m )(130 rpm).
2
When the wheel is inverted, the student and turrable acquire anugular momentum,
L z such that
L z − L0 = L0 or L z = 2 L0 . If I tot is the total moment of inertia about the z-axis,
L z = I tot (70 rpm) = 2(0.22 kg ⋅ m 2 )(130 rpm), or I tot = 0.817 kg ⋅ m 2 . Now I tot
includes the inertia of the turntable, 0.31 kg m2, the student (a cylinder0, (1/2)M(0.5 m)2, and CM of the
wheel, mh2 9which can be neglected if the axes of the wheel and turntable coincide, and the spin axis is
frictionless). Then (1/2)M(0.15 m)2=(0.817-0.31) kg m2, or M=45.1 kg.
(b) The work done by the student’s muscle equals the change in kinetic energy,
Wnc = ∆K = (1 / 2) I tot ω 2 (the kinetic energy of the spinning wheel is unchanged, as is the gravitational
potential energy.) Therefore: Wnc = (1 / 2)(0.817 kg ⋅ m )(70π / 30 s) = 22.0 J.
2
2
Problem 11.55
Description: A thin rod of length Land mass M is free to pivot about one end..
Part A
If it makes an angle with the horizontal, find the torque due to gravity about the pivot point. (You'll
need to integrate the torques on the individual mass elements composing the rod.)
Part B
Find the value of such that when the rod is released from rest at this angle, the vertical component of
the initial acceleration of the rod's tip is equal to (the acceleration of gravity).
Solutions
(a) The rod has mass M and length L, and pivots at one end. The angle it makes with the horizontal is θ.
We use an integral for the torque, although we might expect that this torque was equal to the torque acting
r
r
r
r
on the center of mass of the rod. The torque dτ on a mass element dm is dτ = r &times; dF , so
dτ = xgdm cos θ . We integrate this over the length of the rod, where dm = ( M / L ) dx.
Therefore τ =
∫
L
0
xg cos θ
M
Mg cos θ
dx =
L
L
∫
L
0
xdx =
Mg cos θ ⎛ 1 2 ⎞ 1
⎜ L ⎟ = mgL cos θ .
L
⎝2 ⎠ 2
(b) Since τ
= Iα and a = Lα = g , and the moment of inertia of a rod rotating around one end is
(1 / 2)mgL cos θ 3 g cos θ
I = (1 / 3) ML2 , α = τ / I =
=
. Setting a = Lα = g ,
2 L
(1 / 3) ML2
g = (3 / 2) g cos θ , so θ = cos −1 ( 2 / 3) = 48.2&deg;.
Problem 11.53
Description: A solid ball of mass and radius is spinning with angular velocity about a
horizontal axis. It drops vertically onto a surface where the coefficient of kinetic friction with the ball is
(see the figure). Find expressions for (a) the final angular velocity once it's achieved pure rolling
motion and (b) the time it takes until it's in pure rolling motion.
Solutions
A solid spinning ball drops onto a frictional surface. At first it slides, but due to friction it will slow down
its spin and increase its linear motion until it is purely rolling without sliding. The frictional force is
F f = &micro;Fn = &micro;Mg . There is a torque on the ball due to frictional force, acting on the edge. This torque
τ = − &micro;mgR serves to slow the ball’s rotation: we can use τ = Iα to find the angular acceleration α
and then use ω = ω 0 + αt to find the resulting angular speed. The frictional force on the ball also
accelerates the ball, so we can use f = Ma and v = v 0 + at to find the speed of the ball.
α = τ / I = (− &micro;MgR) /[(2 / 5) MR 2 ] = −5&micro;g /(2 R), so ω = ω 0 − 5&micro;g /(2 R)t. Since
a = F / M = &micro;Mg / M = &micro;g , v = v 0 + &micro;gt. We set
Rω = v = Rω 0 − (5 / 2) &micro;gt = v0 (= 0) + &micro;gt , so solving this for t gives Rω 0 = &micro;gt (1 + 5 / 2).
So t = 7 Rω 0 /( &micro;g ), and plugging this into the equation for ω gives
ω = ω 0 − [5&micro;g /(2 R)][2 Rω 0 /( &micro;g )] = ω 0 − (5 / 7)ω 0 = (2 / 7)ω 0 .
(a) ω = ( 2 / 7)ω 0
(b) t = (7 / 2)[ Rω 0 /( &micro;g )].
Problem 11.38
Description: The figure shows the dimensions of a 880
about its center of mass is 0.048
Part A
wooden baseball bat whose rotational inertia
.
If the bat is swung so its far end moves at 50
, find its angular momentum about the pivot point
. Hint: Remember the parallel-axis theorem.
Part B
Find the constant torque applied about to achieve this angular momentum in 0.25 .
Solutions
(a) The rotational inertia of the bat about point P is
I P = Mh 2 + I cm = (0.88 kg)(0.43 m) 2 + 0.048 kg ⋅ m 2 = 2.11 &times; 10 −1 kg ⋅ m 2 (see Eq.10.17.)
Its angular velocity (direction parallel to the axis of swing) about P is
ω P = v / r = (50 m/s)/(0.74 m) = 67.6 s -1 . Therefore, the angular momentum about P (also
directed along the axis of swing) is LP = I Pω P = 14.2 J ⋅ s.
(b) For constant torque along the axis of swing, τ = ∆L / ∆t implies
τ = (14.2 J ⋅ s)/(0.25 s) = 56.9 N ⋅ m. (if we assume that the bat is initially at rest.)
Problem 11.48
Description: A uniform, solid, spherical asteroid with mass
rotating with a period of 4.3
equator at 8.4
is
. A meteoroid moving in the asteroid's equatorial plane crashes into the
. It hits at a 58 angle to the vertical and embeds itself at the surface. After the
impact the asteroid's rotation period is 3.9 .
Part A
Find the meteoroid's mass.
Solution
If we neglect external gravitational torque, the total angular momentum of the asteroid
and meteoroid is conserved. Since the meteoroid is moving in the asteroid’s equatorial
plane, its angular momentum about the center of the asteroid is in the same direction as
the rotational angular momentum of the asteroid.
r
r
Li = Iω i + R &times; mv = (2 / 5) MR 2ω i + mvR sin 58&deg; and
L f = ( I + mR 2 )ω f = (2M / 5 + m) R 2ω f . Setting Li = L f , solving for m, and using the
period of rotation instead of the angular speed, we find:
0.4 MR (T f−1 − Ti −1 )
m=
= 2.81 &times; 10 7 kg.
−1
vR sin 58&deg; − RT f
```