Question

A cart consists of a body and four wheels on frictionless axles. The body has a mass m. The wheels
are uniform disks of mass M and radius R. The cart rolls, without slipping, back and forth on a
horizontal plane under the influence of a spring attached to one end of the cart (figure). The spring
cơhstånt is k. Taking into account the moment of inertia of the wheels, find a formula for the
frequency of the back and forth motion of the cart.
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Answer 1

Answer:

Explanation:

Let the displacement of cart is x

compression inspring is x

Total energy  E= constant

1/2 kx² + 1/2 mv² + 4 ( 1/2 mv² + 1/2 Iw²) = const

Rolling on ground  v = RW

w = V/R

I = MR² /2

1/2 kx² + 1/2 mv² + 4 ( 1/2 mv² + 1/2 MR² × v²/R²) = const

kx² + mv² + 4 ( 3/2 mv² ) = const

Kx² = - mv² - 6mv²

Kx² = - (m + 6m)v²

K2x× dx/df = -(m + 6m) 2v× dv/df

Kx  = - (m + 6m ) a

x/a = -( m + 6m /k )

T = 2∧ √ m + 6m /k

f = 1/2∧ √ k /m +6m

Answer 2

Given that,

Mass of body = m

Mass of wheels = M

Radius = R

Spring constant = k

Let the displacement of cart is x.

We need to calculate the displacement and acceleration

Using total energy is constant

Total energy = constant

$\dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2+4(\dfrac{1}{2}Mv^2+\dfrac{1}{2}I\omega^2)=constant$

We know that,

The angular velocity is

$\omega=\dfrac{v}{R}$

The moment of inertial is

$I=\dfrac{1}{2}MR^2$

Put the value of angular velocity and moment of inertia in equation (I)

$\dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2+4(\dfrac{1}{2}Mv^2+\dfrac{1}{2}\times\dfrac{1}{2}MR^2\times\dfrac{v^2}{R^2})=constant$

$kx^2+mv^2+4(\dfrac{3}{2}Mv^2)=constant$

$kx^2=-mv^2-6 Mv^2$

$kx^2=-(m+6M)v^2$

On differentiating

$2kx\dfrac{dx}{dt}=-(m+6M)\times2v\dfrac{dv}{dt}$

We know that,

The acceleration is

$\dfrac{dv}{dt}=a$

The velocity is

$\dfrac{dx}{dt}=v$

So, $2kx\times v=-(m+6M)a\times 2v$

$kx=-(m+6M)a$

$\dfrac{x}{a}=-\dfrac{m+6M}{k}$

We need to calculate the time period of the cart

Using formula of time period

$T=2\pi\sqrt{\dfrac{displacement}{acceleration}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{m+6M}{k}}$

We need to calculate the frequency of the cart

Using formula  of frequency

$f=\dfrac{1}{T}$

Put the value into the formula

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m+6M}}$

Hence, The frequency of the cart is $\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m+6M}}$