A cart consists of a body and four wheels on frictionless axles. The b

1.	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 car (figure). The spring constant 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.
Answer» Compression in spring is x, Total energy E = constant \(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{2}Iw^2\)) = constant Rolling on ground v = Rω ω = \(\frac{V}{R}\) ∵ I = \(\frac{MR^2}{2}\) \(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{4}MR^2\) x \(\frac{V^2}{R^2}\)) = constant kx² + mv² + 4(Mv² + \(\frac{1}{2}\)Mv²) = constant kx² + mv² + 4(\(\frac{3}{2}\)Mv²) = constant kx² + mv² + 6Mv² kx² = - (m + 6M)v² .........(1) Different equation (1) k 2x(\(\frac{dx}{dt}\)) = - (m + 6M) 2v\(\frac{dv}{dt}\) kx = -(m + 6M)a x = \(\frac{-(m+6M)a}{k}\) a = \(\frac{-k}{(m+6M)}x\) .........(2) a = - ω²x ..........(3) Compression this equation (2) and (3) Then, ω²= \(\frac{k}{m+6M}\) ⇒ ω = \(\sqrt{\frac{k}{m+6M}}\) Frequency formula, f = \(\frac{\omega}{2\pi}\) ∴ f = \(\frac{1}{2\pi}\)\(\sqrt{\frac{k}{m+6M}}\)

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 car (figure). The spring constant 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.

Answer»

Compression in spring is x,

Total energy E = constant

\(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{2}Iw^2\)) = constant

Rolling on ground v = Rω

ω = \(\frac{V}{R}\)

∵ I = \(\frac{MR^2}{2}\)

\(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{4}MR^2\) x \(\frac{V^2}{R^2}\)) = constant

kx² + mv² + 4(Mv² + \(\frac{1}{2}\)Mv²) = constant

kx² + mv² + 4(\(\frac{3}{2}\)Mv²) = constant

kx² + mv² + 6Mv²

kx² = - (m + 6M)v² .........(1)

Different equation (1)

k 2x(\(\frac{dx}{dt}\)) = - (m + 6M) 2v\(\frac{dv}{dt}\)

kx = -(m + 6M)a

x = \(\frac{-(m+6M)a}{k}\)

a = \(\frac{-k}{(m+6M)}x\) .........(2)

a = - ω²x ..........(3)

Compression this equation (2) and (3)

Then,

ω²= \(\frac{k}{m+6M}\)

⇒ ω = \(\sqrt{\frac{k}{m+6M}}\)

Frequency formula,

f = \(\frac{\omega}{2\pi}\)

∴ f = \(\frac{1}{2\pi}\)\(\sqrt{\frac{k}{m+6M}}\)