Physics Rotational Inertia Help?

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2017-02-06 2:57 am

2 rods with mass M and length L form the cross pieces to a hoop of mass M and radius L/2 to make a wheel. A mass M (same as rods and hoop) is attached to the wheel by a massless cord wrapped around the rim of the wheel (cord and weight hang straight down from a point on the rim horizontal to the left to the axis of the wheel). Write an expression for the tension and acceleration of the mass as it falls.

回答 (1)

NCS

2017-02-06 4:59 am

✔ 最佳答案

moment of inertia of wheel:
I = 2*ML²/12 + M(L/2)² = ML²(1/6 + 1/4) = 5ML²/12 = (5/3)M(L/2)²

shortcut:
motive force F = M*g
equilvalent mass m = M + (5/3)M = (8/3)M
acceleration a = F / m = g / (8/3) = 3g/8 ◄ acceleration
tension T = M(g - a) = M(g - 3g/8) = 5Mg / 8 ◄ tension

long cut:
tension T = M(g - a)
and also T = torque / radius = τ / (L/2)
but τ = I*α = I*a/r = (5/3)M(L/2)² * a / (L/2) = 5MLa / 6, so
T = M(g - a) = (5MLa/6) / (L/2) → mass M cancels
g - a = 5a / 3
g = 8a/3
a = 3g/8 ◄
and T = M(g - 3g/8) = 5Mg / 8 ◄
as before.

Hope this helps!

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