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2 years ago

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A student decides to give his bicycle a tune up. He flips it upside down (so there’s no friction with the ground) and applies a

force of 20 N over 1.2 seconds to the pedal, which has a length of 16.5 cm. If the back wheel has a radius of 33.0 cm and moment of inertia of 1200 kg·cm^2, what is the tangential velocity of the rim of the back wheel in m/s? Assume he rides a fixed gear bicycle so that one revolution of the pedal is equal to one revolution of the tire. Round your answer to 1 decimal place for entry into eCampus. Do not enter units. Example: 12.3

1 answer:

Alinara [238K]2 years ago

7 0

Answer:

Tangential velocity = 10.9 m/S

Explanation:

As per the data given in the question,

Force = 20 N

Time = 1.2 S

Length = 16.5 cm

Radius = 33.0 cm

Moment of inertia = 1200 kg.cm^2 = 1200 × 10^(-4) kg.m^2

= 1200 × 10^(-2) m^2

Revolution of the pedal ÷ revolution of wheel = 1

Torque on the pedal = Force × Length

= 20 × 16.5 10^(-2)

= 3.30 N m

So, Angular acceleration = Torque ÷ Moment of inertia

= 3.30 ÷ 12 × 10^(-2)

= 27.50 rad ÷ S^2

Since wheel started rotating from rest, so initial angular velocity = 0 rad/S

Now, Angular velocity = Initial angular velocity + Angular Acceleration × Time

= 0 + 27.50 × 1.2

= 33 rad/S

Hence, Tangential velocity = Angular velocity × Radius

= 33 × 33 × 10^(-2)

= 10.9 m/S

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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2 years ago

A machine has an efficiency of 20 percent. Find the input work if the output work is 140 Joules

Lapatulllka [165]

Answer:

X= 700 Joules

Explanation:

The question asked about the efficiency of the work done.

The formula for efficiency is: Efficiency = (Useful output / input work) * 100%

The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.

Using the formula of efficiency,

20 = (140/X) * 100

So, we simply solve the above equation.

X= 140*100/20

X= 700 Joules

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Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5

allochka39001 [22]

There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.

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2 years ago

An organ pipe is made to play a low note at 27.5 Hz, the same as the lowest note on a piano. Assuming a sound speed of 343 m/s,

timama [110]

Answer:

The length of open-open pipe needed is 6.23 m

The length of open-close pipe needed is 3.11 m

Explanation:

Fundamental frequency for standing wave mode of an open- open pipe is given by

$f=\frac{v}{2L}$

where v is the velocity and L is the length

The length of open-open pipe needed is

$L=\frac{v}{2f} \\L=\frac{343}{2\times 27.5} \\L=6.23 m$

Fundamental frequency for standing wave mode of an open- close pipe is given by

$f=\frac{v}{2L}$

The length of open-close pipe needed is

$L=\frac{v}{2f} \\L=\frac{343}{2\times 27.5} \\L=6.23 m$

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A 2.0-kg box and a 3.0-kg box on a perfectly smooth horizontal floor have a spring of force constant 250 N/m compressed between

svp [43]

Answer:

7.5 m/s² and 5 m/s²

Explanation:

Note: The force in the compressed spring is the same as the force needed to produce the acceleration in each box.

From Hook's law,

F = ke..................... Equation 1

Where F = force, k = spring constant, e = compression.

Given: k = 250 N/m, e = 6.0 cm = 0.06 m

Substitute into equation 1

F = 250×0.06

F = 15 N.

Using Newton's fundamental equation of kinematics

F = ma.................. Equation 2

Where m = mass of the box, a = acceleration of the box.

make a the subject of the equation

a = F/m................. Equation 3

For the first box,

Given: m = 2.0 kg, F = 15 N

Substitute into equation 3

a = 15/2.0

a = 7.5 m/s².

For the second box,

Give: m = 3.0 kg, F = 15 N

Substitute into equation 3

a = 15/3

a = 5 m/s²

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2 years ago

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