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3 years ago

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A bicycle wheel of radius 14 cm is mounted at the middle of an axle 96 cm long. The tire and rim weigh 23 N. The wheel is spun a

t 12 rev/s, and the axle is then placed in a horizonal position with one end resting on a pivot
a. What is the magnitude of the angular momentum due to the spinning of the wheel? (Treat the wheel as a hoop.)
b. What is the angular speed of precession? rad/s How long does it take for the axle to swing through 360 degree around the pivot?
c. What is the magnitude of the angular momentum associated with the motion of the center of mass, that is, due to the d. precession? In what direction is this angular momentum? up up or down, depending on the direction of in part (a) down

1 answer:

ollegr [7]3 years ago

8 0

Answer:

a. L =3.465 J.s

b. 3.186 rad/s

c. T = 1.972 s

d. Lp = 0.022 J.s; up or down, depending on the direction of L

Explanation:

(a)We have the formula for angular momentum as

L = I*w where I = moment of inertia = $M*R^2$

L = $(23/9.81) * 0.14^2* (24*pi) J.s = 3.465 J.s$

(b) ω = M*g*D/L

where D = L/2 = 0.96/2 = 0.48m

ω = (23* 0.48/3.465) rad/s = 3.186 rad/s

(c) T = 2π/ω

T = 2*π/3.186 s = 1.972 s

(d) Lp = $M*D^(2*w)$

Solving with mass M, D and angular velocity ω

Lp = 0.022 J.s; up or down, depending on the direction of L

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1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

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$S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m$

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$S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s$

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100mL of 4°C water is heated to 37 °C . Assume the density of the water is 1g/mL. The specific heat of water is 4.18 J/g(°C). Wh

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

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Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

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The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

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