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julia-pushkina [17]

2 years ago

8

A flywheel in the form of a uniformly thick disk of radius 1.33 m1.33 m has a mass of 70.6 kg70.6 kg and spins counterclockwise

at 217 rpm217 rpm . Calculate the constant torque required to stop it in 2.75 min2.75 min .

1 answer:

ladessa [460]2 years ago

3 0

Answer:

The constant torque required to stop the disk is 8.6 N-m in clockwise direction .

Explanation:

Let counterclockwise be positive direction and clockwise be negative direction .

Given

Radius of disk , r = 1.33 m

Mass of disc , m = 70.6 kg

Initial angular velocity , $\omega_i =217 rpm$

Final angular velocity , $\omega_f =0\, rpm$

Time taken to stop , t = 2.75 min

Let $\alpha$ be the angular acceleration

We know

$\omega _f=\omega _i+\alpha t$

=> $0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}$

=> $\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}$

Torque required to stop is given by

$\tau =I\alpha$

where moment of inertia , $I=\frac{mr^{2}}{2}=\frac{70.6\times 1.33^{2}}{2}kg.m^{2}=62.5 kg.m^{2}$

=> $\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m$

Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .

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3 years ago

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3 years ago

Read 2 more answers

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$

Potential energy is given by mgh

According to energy conservation

$mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2$

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3 years ago

Assuming the pick-up trucks, trailers and road conditions are exactly the same, which vehicle will take a longer distance to sto

mars1129 [50]

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6 0

3 years ago

A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

matrenka [14]

Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

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3 years ago