Ch 10.1 #1

During a certain period of time, the angular position of a swinging door is described by θ = 2t2 + 10t + 5, where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

q = 5 + 10t + 2t2

w = dq/dt

w = 10 + 4t

a = dw/dt

a = 4

(a) t = 0

q = 5 rad

w = 10 rad/s

a = 4 rad/s2

(b) t = 3

q = 53 rad

w = 22 rad/s

a = 4 rad/s2

 

 

 

Ch 10.2 #2

A dentists drill starts from rest. After 3.2 s of constant angular acceleration, it turns as a rate of 2.51 x 104 rpm. (a) Find the drills angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period. a = Dv/Dt a = Dw/Dt

(a) a = 2.51 x 104 rpm (2prad/rev) (1min/60s)/ 3.2 sec

a = 821 rad / s2

(b) q = a t2

q = 4200 radians

 

 

 

Ch 10.2 #8

A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00 s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

a = (ωf ω0) / t

ω0 = ωf - a t

OR

ωave = Δq / Δ t

f + ω0) = (qf - q0) / t

(98 + ω0) = 37(2 π) / 3

ω0 = 57 rad

q - q0 = at2 + f - a t)t

q = ωft - at2

37(2π) = 98(3) a32

a = 13.7 rad/s2

a = (ωf ω0) / t

a = (98 57) / 3

a = 13.7 rad/s2

 

 

Ch 10.3 #17

A disk 8 cm in radius rotates at a constant rate of 1200 rpm about its central axis. Determine (a) its angular speed, (b) the tangential speed at a point 3 cm from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2 sec.

(a)

w = 1200 rpm (2prad/rev) (1min/60s)

w = 126 rad / sec

(b) v = w r

v = 126rad/sec (0.03m)

v = 3.77 m/s

(c) a = v2 / r = w2r2 / r

a = w2r

a = 1260 m/s2

 

 

Ch 10.4 #21

The 4 particles are connected by rigid rods of negligible mass. The origin is at the center. If the system rotates in the x-y plane about the z axis at 6 rad/sec, calc

(a) moment of inertia of the system about the z-axis

(b) the rotational kinetic energy of the system

r1 = r2 = r3 = r4

r = ((6/2)2 + (4/2)2)

r = 13 meters

 

(a) I = Sjmjrj2

I = (3 + 2 + 2 + 4) 13

I = 143 kg m2

(b)

KR = Iw2

KR = 143 (6 rad/s)2

KR = 2.57 x 103 J

 

 

 

Ch 10.5 #23

Three identical this rods, each of length L and mass m are welded perpendicular to one another. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. Determine the moment of inertia of this structure.

Ans: 11 mL2 / 12

 

Hint:

We need to add the Inertia for all three rods

I = Ix + Iy + Iz

untitled1

 

For Iy all of the mass is located at L/2 from the pivot point

Iy = m(L/2)2

Iy = 1/4 mL2

 

For Iz the rod on the z-axis is being pivoted about the end (formula from chart)

Iz = 1/3 mL2

 

For Ix the mass varies from L/2 to L/√2

The Right Half (also can be done all at once)

Ix = r2dm l = m/L and l = dm/dx

Ix = r2 l dx where r2 = x2 + (L/2)2

And x varies from 0 to L/2

Work to the right

 

I = Ix + Iy + Iz

I = mL2/3 + mL2/4 + mL2/3

I = 11mL2/12

untitled2

Right Half of x

Ix = r2 l dx where r2 = x2 + (L/2)2

Ix =l (x2 + (L/2)2) dx

Ix = l x2dx + l (L/2)2 dx (x 0 to L/2)

Ix = l/3 x3 + l/4 L2 x (x 0 to L/2)

Ix = l/3 (L3/8) + l/4 L2 (L/2)

Ix = (m/L)(L3/24) + (m/L) (L3/8)

Ix = mL2/24 + mL2/8

Ix = mL2/6 (For right half)

Ix = mL2/3 for all of the rod on the x-axis

 

 

 

Ch 10.6 #31

Find the net torque on the wheel about the axle through O if a = 10 cm and b = 25 cm.

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S t = tccw - t cw

S t = 0.1m*12N 0.25m*(9N + 10N)

S t = -3.55 Nm

 

 

 

Ch 10.7 #35

A model airplane with mass 0.75 kg is tethered by a wire so that it flies in a circle 30 m in radius. The airplane engine provides a net thrust of 0.8 N perpendicular to the tethering wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in level flight. (c) Find the linear acceleration of the airplane tangent to its flight path.

(a) t = r x F

t = 30 m * 0.8 N

t = 24 Nm

(b) t = I a

24 = (0.75 * 302) a

a = 0.0356 rad / s2

(c) aT = a * r

aT = 0.0356 * 30

aT = 1.07 m/s2

 

 

 

Ch 10.8 #47 This problem describes an experimental method for determining the moment of inertia of an irregular shaped object such as the payload for a satellite. A counter weight of mass m, is suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. The turntable can rotate w/o friction.

When the counterweight is released from rest it descends through a distance h, acquiring a speed v.

Show that the moment of inertia I of the rotating apparatus is

mr2 (2gh/v2 1).

untitled8

 

KR + K = Ugrav

Iw2 + mv2 = mgh

Iw2 = mgh - mv2

I = 2(mgh - mv2) / w2

I = (2mgh - mv2) / (v/r)2

I = mr2 (2gh - v2) / v2

I = mr2 (2gh/v2 - 1)

 

 

 

 

Ch 10.9 #52

A bowling ball has mass M, radius R, and a moment of inertia of 2/5 MR2. If it starts from rest, how much work must be done on it to set it rolling without slipping at a linear speed, v? Express the work in terms of M and v.

Work = DK

Work = K + KR

Work = Mv2 + Iw2

Work = Mv2 + 2/5MR2 (v/R) 2

Work = Mv2 + 2/10 Mv2

Work = 7/10 Mv2