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3 years ago

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The 10-kg double wheel with radius of gyration of 125mm about o is connected to the spring of stiffness k = 600 n/m by a cord wh

ich is wrapped securely around the inner hub. if the wheel is released from rest on the incline with the spring stretched 225 mm, calculate the maximum velocity v of its center o during the ensuing motion. the wheel rolls without slipping

1 answer:

morpeh [17]3 years ago

7 0

Answer:

Explanation:

Here, spring constant is (k) and the inclination angle of the surface is θ

The initial and final positions of the double wheel are considered. Then change in gravitational potential energy, change in kinetic energy, and the change in elastic potential energy of spring are calculated.

Calculate the change in total kinetic energy

$\Delta T=\frac{1}{2} mv^2+\frac{1}{2} Iw^2\\\\=\frac{1}{2} mv^2+\frac{1}{2} (mk^2)(\frac{v}{r} )^2$

Here, mass of the double wheel is m

velocity of the wheel is v

mass moment of inertia of the wheel is I

angular velocity of the wheel is ω

the radius of gyration of the wheel is k

Substitute 10kg for m, 0.125 for k and 0.2 for r

$\Delta T =\frac{1}{2} (10kg)v^2+\frac{1}{2} *10kg*(0.125)^2*(\frac{v}{0.2} )^2\\\\=6.953v^2$

Determine the inclination of the surface with horizontal.

$\theta = \tan ^-^1(\frac{1}{5} )\\\\=11.31^\circ$

Calculate the change in gravitational potential energy

$\Delta V= mg(x \sin \theta)$

Here, acceleration due to gravity is g

and the displacement of wheel in horizontal direction is x

Substitute 10kg for m, 9.81m/s² for g, 11.31° for θ

$\Delta V= 10*9.81*(x \sin 11.31^\circ)\\\\=19.23x$

The angle rotated by the wheel α to move x distance is obtained as

$x=200\alpha$

$\alpha =\frac{x}{200}$

Calculate the stretch in the spring

$s = (200+75)\alpha$

substitute x/200 for α

$s=(200+75)\frac{x}{200} \\\\=1.375x$

Calculate the final stretch in the spring.

$x_2=0.225-s$

substitute 1.375x for s

$x_2=0.2225-1.375x$

Calculate the change in the elastic potential energy of the spring

$\Delta V_e = \frac{1}{2} k(x_2^2-x^2_1)$

Substitute 600N/m for k, 0.225 for x₁ and (0.225 - 1.375x) for x₂

$\Delta V_e = \frac{1}{2} *600((0.225-1.375x)^2-(0.225)^2)\\\\=567.18x^2-185.62x$

Write the conservation of energy equation for the initial and final positions of the wheel.

$\Delta T+\Delta V+\Delta V_e = 0$

Substitute 6.953v² for ΔT, 19.23x for ΔV and (567.18x² - 185.62x) for ΔV_e

$6.953v^2+19.23x+(567.18x^2-185.62x)=0\\\\\v^2=-81.573x62+23.931x---(1)$ ..........(1)

Calculate the value of x for which maximum velocity is obtained by differentiating

$\frac{d}{dx} (v) = 0$

Differentiate Equation (1) with respect to x.

$2v\frac{dv}{dx} =-(2 \times 81.573)x+23.931$

Substitute 0 for dv/dx

$2v(0)=-(2*81.573)x+23.931\\\x=0.1467m$

Substitute 0.1467 m in Equation (1)

$v^2=-81.573(0.1467m)^2+23.931(0.1467m)\\\\v=1.325m/s$

The maximum velocity of the wheel’s center O is 1.325 m/s

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