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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago

5 meddi muttasıl örnek

alexandr1967 [171]

Answer:

WHAT'S YOUR LANGUAGE

I CAN'T UNDERSTAND

5 0

2 years ago

Read 2 more answers

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$