Mechanics

LUMS School Of Science And Engineering PH-101

Solution Set # 8

Fall 2011-12

December, 2, 2011

Answer 1: A hard-boiled egg spins faster than a raw-egg. The reason for this has to do with the inside. The hard-boiled egg is rigid, while raw-egg is non-rigid because it has liquid inside. When spin a hard-boiled egg all of the contents are spin together, so all the force applied is transferred to motion at the same rotational rate and support keeping it spinning. When we spin a raw-egg the solid shell is moving full speed at the start, but it has to drive the liquid inside in shear, and closer we get to the middle, the more the liquid slips behind to slip in the viscous liquid. The result is that we don’t get the inside moving up to speed and as soon as we let goof the shell, the still slower moving insides start to drag on the shell and slow it down. In other words we can say that the work involved in case of turning the hard-boiled egg is much greater than in case of the raw one and, in consequence, the energy of the rotational motion is much greater than the one acquired by the raw egg. However, if we turned the raw egg long enough we could make it spin just as well as the hard-boiled one. Interestingly, since it may be hard to judge speeds, the quickest way to tell if an egg is raw or boiled is to spin it and then tap it, to bring it to a stop. If the egg is raw, the moving inside will start it moving again, while a hard-boiled egg will just stay stopped. Answer 2: Because of the angular momentum of the spinning horizontal main rotor, the big one above the helicopter, the equal and opposite reaction would be below the rotor to conserve the angular momentum. To counter balance this, we have a small vertical propeller at the tail of the helicopter, which creates a thrust opposite to the rotational force exerted by the main rotor above, so that helicopter can stay facing the desired direction. The helicopter turn right and left by increasing or decreasing the force produced by the vertical propeller. If you watch enough action movies, sometime they shoot out the small rotor and helicopter spins out of control. That’s pretty much what would happen without it. Answer 3: Let us suppose that hula hoop of mass m and radius r rolls down a slope of 1 : 10 without slipping, as shown in the figure below.

1

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

ω

l

h=1

θ

x=10

1 Linear kinetic energy of the hoop = K.Elinear = mv 2 2 1 Rotational kinetic energy of the hoop = K.Erotational = Iω 2 2 Where I is the moment of inertia of hoop and ω is its angular velocity. Since hoop is rolling down without slipping, therefore ω = v/r. Namely we know that, moment of inertia of hoop of mass m and radius r is, I = mr2 Then rotational kinetic energy will become, K.Erotational

( )2 1 2 v = mr 2 r 1 2 = mv 2

Now according to law of conservation of energy, Potential energy at the top = Kinetic energy at the bottom 1 1 mgh = mv 2 + Iω 2 2 2 1 2 1 2 mgh = mv + mv 2 2 2 mgh = mv v 2 = gh

(1)

From the figure it is clear that, h = ℓ sin θ 2

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

Substitute value of h in equation (1), we get, v 2 = gℓ sin θ

(2)

Where ℓ is the length of the slope. Since height of the slope is very less that the length of the slope. Final velocity of the hoop at the bottom of the slope can be calculated by the use of third equation of motion, i.e. 2as = vf2 − vi2 2aℓ = v 2 − 0 v 2 = 2aℓ

(3)

Comparing equation (2) and (3), we get 2aℓ = gℓ sin θ 1 a = g sin θ 2 Last expression gives the required acceleration. This equation shows that the linear acceleration of the hoop is independent of the mass of the hoop and length of the roll. It depends solely on g, which is fixed and the angle of the ramp. From figure it is clear that, tan θ =

1 10

θ = tan−1

1 10

= 5.71o Also we know that g = 9.8 m/s2 , therefore numerical value of linear acceleration will be, 1 × 9.8 × sin(5.71) 2 = 0.49 m/s2

a =

Answer 4: We have given that, Mass of meteoroid = m = 150kg Speed of meteoroid = v = 60 km/s = 60, 000 m/s Distance of meteoroid from centre of earth = r = 1.2 × 104 km = 1.2 × 107 m 3

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

a). Angular momentum of meteoroid will be, Lmeteoroid = mvr = 150 × 60, 000 × 1.2 × 107 = 1.08 × 1014 kgm2 /s b). We know that mass of Earth is M = 6.0 × 1024 kg, angular momentum of Earth will be, Learth = M vr = 6.0 × 1024 × 60, 000 × 1.2 × 107 = 4.3 × 1036 kgm2 /s Answer 5: ω1 R2

R1

ω2

a). Following information for the two flywheels are given. M1 = Mass of first flywheel R1 = Radius of first flywheel ω = Initial angular velocity of first flywheel ω1 = Final angular velocity of first flywheel M2 = Mass of second flywheel R2 = Radius of second flywheel 0 = Initial angular velocity of second flywheel ω2 = Final angular velocity of second flywheel

4

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

Since after coupling both the flywheels have same velocity, therefore, v1 = v2 R1 ω1 = R2 ω2

(4)

Apply law of conservation of angular momentum, according to which,     Total angular momentum Total angular momentum   =   before coupling after coupling (L1 + L2 )initial = (L1 + L2 )final (I1 ω + 0) = (I1 ω1 + I2 ω2 ) Since both flywheel are disks, 1 M1 R12 2 1 I2 = M2 R22 2 I1 =

Therefore, 1 1 1 M1 R12 ω = M1 R12 ω1 + M2 R22 ω2 2 2 2 2 2 M1 R1 ω = M1 R1 ω1 + M2 R22 ω2 R1 ω1 M1 R12 ω = M1 R12 ω1 + M2 R22 · R2 M1 R1 ω = M1 R1 ω1 + M2 R2 ω1 M1 R1 ω ω1 = (M1 R1 + M2 R2 )

∵ R1 ω1 = R2 ω2

Substitution of the value of ω1 in equation (4) yields, ω2 =

M1 R12 ω (M1 R1 + M2 R2 )R2

b). In order to calculate the loss in kinetic energy, firstly we shall calculate initial and final kinetic energies. Initial kinetic energy is, 1 I1 ω 2 2( ) 1 1 2 M1 R1 ω 2 = 2 2 1 = M1 R12 ω 2 4

Initial K.E =

5

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

While final kinetic energy is, 1 1 I1 ω12 + I2 ω22 2( 2) ( ) 1 1 1 1 2 2 2 M1 R1 ω1 + M2 R2 ω22 = 2 2 2 2 1 1 = M1 R12 ω12 + M2 R22 ω22 4 4

Final K.E =

Substitute values of ω1 and ω2 , we have, )2 )2 ( ( 1 1 M1 R1 ω M1 R12 ω 2 2 Final K.E = M1 R1 + M2 R2 4 (M1 R1 + M2 R2 ) 4 (M1 R1 + M2 R2 )R2 3 4 2 4 2 (M1 R1 + M1 M2 R1 )ω = 4(M1 R1 + M2 R2 )2 Therefore change in kinetic energy will be, Change in K.E = Initial K.E − Final K.E 1 (M13 R14 + M12 M2 R14 )ω 2 = M1 R12 ω 2 − 4 4(M1 R1 + M2 R2 )2 M1 M2 R12 (M2 R22 + 2M1 R1 R2 − M1 R12 )ω 2 = 4(M1 R1 + M2 R2 )2 Loss in kinetic energy is used in work done to overcome force of friction between the surfaces of belt and flywheels. Answer 6:

r

C

M

T T

m mg To solve the problem we first examine the figure to establish a simple sign convention. The weight will be accelerated in the downward direction by the force of gravity and at the same time the disk will be caused to rotate in a clockwise direction. Let us call the clockwise direction the positive direction of rotation and the downward direction the positive direction 6

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

of translation. From the nature of the force which can be exerted by a string, the upward force exerted by the string on the falling mass must be equal in magnitude to the downward force exerted on the disk by the string. We shall call the magnitude of this force T and indicate the directions on the diagram. From the nature of the constraints, we note that the only way the mass can receive a given downward displacement is for a length of string to unroll from the disk. If the radius of the disk is r, the disk must receive an angular displacement θ when the mass is displaced by a distance s such that s = rθ. Note that the directions of the angular and linear displacements as well as their magnitudes are related by this equation, for a positive value of s implies a positive value of θ. Thus we have v = ωr and a = αr, where v and a are the velocity and acceleration of the mass, positive downward and ω and α are the angular velocity and angular acceleration of the disk, positive clockwise. With the relationships of constraint and the sign conventions established, we proceed to a formal solution of the problem. Let us first consider the forces acting on the mass m. From Newton’s second law we can write, mg − T = ma

(5)

Now let us consider the rotation of the disk. The only force acting on the disk that exerts a torque about the axis of rotation through C is the pull of the cord T . If r is the radius of the disk and I is its moment of inertia, we find, τ = T r = Iα and since ⇒ we have

a = αr a α = r Ia Tr = r Ia T = 2 r

Use this value of T in equation (5), we have, mg −

Ia = ma r2

7

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

The moment of inertia of a uniform disk about an axis through C is 21 M r2 . a 1 · M r2 = ma 2 r 2 2mg a= (2m + M )

mg − ⇒ Answer 7:

y

x

l 4

C

l

x

Flat friction-less surface y a). We have given that, Mass of rod = M Length of rod = ℓ Mass of putty = m Initial velocity of putty = v In order to calculate position of mass with adhering putty, let us consider the centre of the rod as origin. When the putty stick the rod at a distance of (ℓ/4) from the centre, the new position of centre of mass can be calculated by the following equation. ∑N mi ri Rcm = ∑i=1 N i=1 mi M (0) + m(ℓ/4) Rcm = M +m mℓ = 4(M + m)

8

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

b). We know that, ∑N Vcm = ∑i=1 N

mi vi

mi M (0) + mv = M +m mv = M +m i=1

c).

y

x

l Rcm 4 C Rcm

l 4

C

x l

Flat friction-less surface y ′

We have given that velocity of putty before collision is v. Let v is the velocity of putty in ′

cm-frame. Then v = v − Vcm . Total angular momentum of the system will be equal to the sum of angular momentum of rod and putty. Therefore, ′

|Lcm | = mv r + M Vcm Rcm = m(v − Vcm )r + M Vcm Rcm

(6)

Where r represents position of putty relative to new centre of mass, i.e. r=

ℓ mℓ − 4 4(M + m)

Substitute values in equation (6), we have, ( )( ) ( )( ) mv mℓ ℓ mv mℓ |Lcm | = m v − − +M (M + m) 4 4(M + m) M +m 4(M + m) 2 2 m M vℓ mM vℓ + = 2 4(M + m) 4(M + m)2 mM vℓ = 4(M + m) 9

Mechanics

LUMS School Of Science And Engineering

Fall 2011-12

Now moment of inertia of the system will be equal to the sum of moment of inertia of ball and rod, i.e. Icm = Irod + Iball ( ) ( )2 1 ℓ 2 2 = M ℓ + M Rcm + m − Rcm 12 4 ( )2 ( )2 mℓ mℓ 1 ℓ 2 = Mℓ + M +m − 12 4(M + m) 4 4(M + m) 1 m 2 M ℓ2 mM 2 ℓ2 = M ℓ2 + + 12 16(M + m)2 16(M + m)2 1 mM ℓ2 = M ℓ2 + 12 16(M + m) 2 M ℓ (4M + 7m) = 48(M + m) Angular frequency can be calculated by the following relation. |Lcm | = Icm ω |Lcm | ⇒ ω = Icm 48(M + m) mM vℓ × = 4(M + m) M ℓ2 (4M + 7m) 12mv = ℓ(4M + 7m)

10