PY1052 Problem Set 4 – Autumn 2004Solutions

(1) An old streetcar rounds a flat corner of radius 9.1 m at a speedof 16 km/hr. What angle with the vertical will be made by the looselyhanging hand straps?

mg

T

to centre of circle

Tcos φ

φ

Tsin φ

The straps hang at an angle φ outward from the vertical. The vertical (y) com-ponent of the tension balances the weight (no acceleration in the vertical direction),while the horizontal (r) component of the tension provides the centripetal force:

y : T cosφ−mg = 0

r : T sinφ =mv2

rSolving for T in the first equation and substituting it into the second yields

mg tanφ =mv2

r

tanφ =v2

rg

Inserting v = 16 km/hr = 4.44 m/s, r = 9.1 m, and g = 9.8 m/s2 yields φ = 12.5◦.

(2) An airplane is flying in a horizontal circle at a speed of 480 km/hr.If its wings are tilted 40◦ to the horizontal, what is the radius of the circlein which the plane is flying? Assume that the required force is providedby a lift force that is perpendicular to the wing surface.

40

The vertical (y) component of the lift force L balances the weight (no accelera-tion in the vertical direction), and the horizontal (r) component of L provides thecentripetal force: y : L cos(40)−mg = 0

r : L sin(40) =mv2

rSolving for L in the first equation and substituting it into the second yields

mg tan(40) =mv2

r

r =v2

g tan(40)

Inserting v = 480 km/hr = 133.3 m/s and g = 9.8 m/s2 yields r = 2161 m.

(3) A student with a mass of 60 kg sits on a Ferris wheel with radius15 m rotating with constant speed v. At the highest point, the magnitudeof the normal force ~N on the student exerted by the seat on which sheis sitting is 460 N. (a) Draw a diagram showing the forces acting on thestudent when she is at the highest point. Does she feel “light” (the normalforce is less than usual) or “heavy” (the normal force is more than usual)at this point? (b) What is the speed with which the Ferris wheel turns?(c) What is the magnitude of the normal force at the lowest point? (d)What is the normal force at the lowest point if the speed of the Ferriswheel is doubled? N

mg

(a) At the top, the normal force of the seat on the student is upward, and her weight(like always) is downward. Because she is moving in a circle, there must be a netforce pointing inward toward the centre of the circle. Therefore, even if we weren’ttold the magnitude of the normal force at the top, we know that this normal force Nmust be smaller in magnitude than her weight mg, so that the sum of the two pointsinward toward the centre of the circle (which is downward at the top point). Thus,N is smaller than its usual value (mg), and she feels light as she goes over the top.

(b) At the top, taking inward to be the negative direction,

N −mg = −mv2

R

v =

√(mg −N)R

m

=

√√√√ [(60 kg)(9.8 m/s2)− 460 N](15 m)

60 kg

= 5.66 m/s

mg

N

(c) At the lowest point, the normal force is again pointing upward, which is nowinward toward the centre of the circle. Thus, again taking inward to be the negativedirection,

mg −N = −mv2

R

N = mg +mv2

R

= (60 kg)(9.8 m/s2) +(60 kg)(5.66 m/s)2

15 m= 716 N

Thus, the student feels “heavy” at the bottom - the normal force is greater than itsusual value of mg = 588 N.

(d) If the speed is doubled, the normal force will be increased to

N = (60 kg)(9.8 m/s2) +(60 kg)(11.32 m/s)2

15 m= 1101 N

(4) A simple pendulum of length 1.2 m is drawn aside an angle of 6.0◦

with the vertical and is then released, so that it executes oscillations in avertical plane. The mass of the bob is 300 g. Determine (a) the period ofthe oscillations, (b) the maximum displacement smax of the bob along thecircular arc along which it travels, (c) the speed of the bob when it passesthrough the central position, (d) the centripetal acceleration of the bob asit passes through the central position, (e) the tension in the string whenthe bob is at the central position, and (f) the tension in the string whenthe bob is at its largest displacement from the vertical.

(a) The period of the oscillations of a simple pendulum is given by

T = 2π√

lg

= 2π√

1.2 m9.8 m/s2 = 2.20 s

(b) The maximum displacement of the bob is the arc length corresponding to theangle θmax = 6.0◦, where theta is given in radians:

θmax = smaxl

smax = θmaxl = [6.0◦ × π180◦ ](1.2 m) = 0.126 m

(c) The speed of the bob when it passes through the central position is its maxi-mum speed, given by

vmax = smaxω

where ω is the angular frequency of the oscillations, which is ω =√g/l:

vmax = smax√

gl

= 0.126 m

√9.8 m/s2

1.2 m= 0.360 m/s

(d) The centripetal acceleration of the bob as it passed through the central positionis

ac = v2

r= (0.360 m/s)2

1.2 m= 0.108 m/s2

(e) At the central position, we have

T −mg = m v2

r−→ T = mg +m v2

r= m(g + v2

r)

T = (0.300 kg)(9.8 m/s2 + 0.108 m/s2) = 2.97 N

(f) At the largest displacement, the tension must be equal to the radial componentof the weight, mg cos(6◦):

T

6 deg

mg

mg cos(6)

Therefore,

T = (0.300 kg)(9.8 m/s2) cos(6) = 2.92 N

(5) An object of mass 0.12 kg is undergoing simple harmonic motionwith an amplitude of 8.5 cm and a frequency f = 5.0 Hz. What are the (a)period and (b) maximum speed of the object? (c) What is the maximumforce it feels during its motion? (d) If the oscillations are produced by aspring, what is the spring constant?

(a) The period of the motion is the inverse of the frequency:

T = 1f

= 15.0 Hz

= 0.20 s

(b) The maximum speed of the object related to the amplitude and angular fre-quency by the expression

vmax = smaxω = smax(2πf) = (0.085 m)2π(5.0 Hz) = 2.67 m/s

(c) The maximum force is equal to the mass times the maximum acceleration,with the maximum acceleration being ω2smax:

Fmax = mamax = mω2smax = m(2πf)2smax = (0.12 kg)(2π(5.0 Hz))2(0.085 m) =10.07 N

(d) If we have a spring,

ω = 2πf =√

km

k = m(2πf)2 = (0.12 kg)(2π(5.0 Hz))2 = 118.4 N/m

(6) A block weighing 18.0 N is resting on a spring at the bottom of africtionless 42◦ ramp. The spring’s unstretched length is 0.45 m and itsspring constant is 100 N/m. (a) How much is the spring compressed bythe block? (b) If the block is pulled slightly up the ramp and released,what is the period of the resulting oscillations?

mg

mg sin(42)42 deg

kx

N

mg cos(42)

(a) In the direction along the ramp, there are two forces acting, which add to zero,since there is no acceleration:

kx−mg sin(42) = 0

x = mg sin(42)k

= (18.0 N) sin(42)100 N/m

= 0.120 m = 12.0 cm

(b) The period of the oscillations is given by the usual expression:

T = 2π√

mk

where the mass of the block is m = 18.0 N/9.8 m/s2 = 1.84 kg:

T = 2π√

1.84 kg100 N/m

= 0.85 s

(7) On August 10, 1972, a large meteorite (m = 4 × 106 kg) skippedacross the atmosphere above the Western United States and Canada. Theaccompanying light given off as the outer parts of the meteorite burnedaway was so bright that it was visible even during the day. If the meteoritehad hit the Earth, its speed on impact would have been 15 km/s. Whatwould have been the meteorite’s kinetic energy upon impact?

The kinetic energy of any moving object is

KE = 12mv2 = 1

2(4× 106 kg)(1.5× 104 m/s)2 = 4.5× 1014 J

(8) Shown below are three forces that are applied to a trunk that movesto the left by 3.00 m over a frictionless floor. The magnitudes of the forcesare F1 = 5.00 N, F2 = 9.00 N and F3 = 3.00 N. During the displacement, whatis the net work done on the trunk by each of the three forces? Does thekinetic energy of the trunk increase or decrease?

60F

F

F

1

2

3In each case, the work done by one of the forces is given by W = Fd cosφ, where

d = 3.00 m and φ is the angle between the force and the displacement.

W1 = (5.00 N)(3.00 m) cos(0) = 15.0 J

W2 = (9.00 N)(3.00 m) cos(120) = −13.5 J

W3 = (3.00 N)(3.00 m) cos(90) = 0 J

The kinetic energy of the trunk increases, since the total amount of work done on itis WT = 15.0− 13.5 + 0 = 1.5 J, which is positive.

(1b) A pail of water is swung in a vertical circle of radius 1.2 m. Whatis the minimum speed at the top of the circle in order for the water tostay in the pail?

In order for the water to stay in the pail, the normal force of the bottom of thepail on the water must be greater than zero. At the top of the circular arc, the normalforce and gravitational force both point downward, and will the sum of these forceswill provide the centripetal force:

N +mg = mv2

r

The higher the speed of the circular motion, the larger the normal force will be.When the speed is such that the normal force is zero, the water is no longer in contactwith the pail – the speed must therefore be greater than this critical value:

mg = mv2c

r−→ vc =

√rg

vc =√

(1.2 m)(9.8 m/s2) = 3.43 m/s

(2b) A 1.50-kg ball is connected by two massless strings to a verticalrotating rod as shown. The strings are tied to the rod and are taut. Thetension in the upper string is 35 N. Find (a) the tension in the lower string,(b) the net force on the ball, and (c) the angular speed of the ball.

(a) Since the three sides of the triangle formed by the two strings and the lengthof the rod between the places they are tied are equal, we know the angles in thistriangle:

60

Tup

Tdown

30

mg

There is no acceleration in the vertical direction, so

Tup,y − Tdown,y −mg = 0

Tup sin(30)− Tdown sin(30)−mg = 0

Tdown = Tup −mg

sin(30)

= 35 N− (1.50 kg)(9.8 m/s2)

sin(30)

Tdown = 5.60 N

(b) The net force on the ball will be the centripetal force, which is provided by thesum of the two horizontal components of the two tensions:

Fc = Tup,r + Tdown,r

= Tup cos(30) + Tdown cos(30)

= (Tup + Tdown) cos(30)

= (35 N + 5.6 N) cos(30) = 35.2 N

(c) The angular speed of the ball can be found from the centripetal force:

Fc =mv2

r= mω2r

ω =

√Fcmr

=

√Fc

ml cos(30)

=

√35.2 N

(1.50 kg)(1.70 m cos(30))

= 3.99 rad/s

(3b) The acceleration of gravity g varies with location on the Earthdue to the Earth’s rotation and the fact that the Earth is not exactlyspherical. This was first discovered in the 17th century, when it was notedthat a pendulum clock carefully adjusted to keep time in Paris lost about90 s/day near the equator.

(a) Show that a small change in g produces a small change in the pen-dulum’s period T given by

∆TT≈ −1

2∆gg

(b) How great a change in g is needed to account for a change in theperiod of 90 s/day?

(a) Treating the length l of the pendulum as a constant and g as a variable, we have

T = 2π

√l

g= 2π

√lg−1/2

dT

dg= 2π

√l[−1

2g−3/2] = − T

2gdT

T= −1

2

dg

g∆T

T≈ −1

2

∆g

g

(b) The fractional change in each period ∆T/T will be equal to the number of secondslost divided by the number of seconds in a day:

∆T

T= − 90 s

86, 400 s≈ −1

2

∆g

g∆g

g≈ 2.08× 10−3

∆g ≈ 2.0× 10−2

(4b) A simple pendulum 0.50 m long is suspended from the roof of arailway cart accelerating in the horizontal direction with a = 7 m/s2. Whatis the period of small oscillations of the pendulum about its equilibriumangle?

mg

T

ma

T

mg

mg

F tan

tan

ma

F = mg sin (φ+θ)

ma cos (φ+θ)

φφ θ

The equilibrium position of the pendulum bob in the accelerating train is givenby (diagram to the left)

~T + ~mg = m~a

tanφ = ag

= 7.09.8−→ φ = 35.5◦

If the bob is pulled aside a small angle θ, the force in the tangential direction willbe mg sin(φ+ θ). This will be equal to the restoring force ms̈ plus the component ofthe train’s acceleration in this direction, which is ma cos(φ+ θ):.

mg sin(φ+ θ) = ms̈+ma cos(φ+ θ)

g[sinφ cos θ + cosφ sin θ] = s̈+ a[cosφ cos θ − sinφ sin θ]

Making the approximation that θ is small, so that sin θ ≈ θ and cos θ ≈ 1, andalso substituting θ = s/l,

g(sinφ+ cosφθ) = s̈+ a(cosφ− sinφθ)

(g cosφ+ a sinφ)s

l= s̈+ (a cosφ− g sinφ)

(g cosφ+ a sinφ)s

l= s̈

(g cosφ+ (g tanφ) sinφ)s

l= s̈

(g cosφ+ (gsinφ

cosφ) sinφ)

s

l= s̈

g

l cosφ(cos2 φ+ sin2 φ)s = s̈

g

l cosφs = s̈

where we have also used the fact that tanφ = ag, so that g sinφ = a cosφ and

a = g tanφ.

We can see that this is the equation for simple harmonic motion, with the expressiong

l cosφin place of the usual g

l. Thus, the period of the motion will be

T = 2π√

lg−→ 2π

√l cosφg

T = 2π√

(0.5 m) cos(35.5)

9.8 m/s2 = 1.28 s