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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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Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2} }{E_{v} ^{2} +(2.3E_{v} )^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$

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3 years ago