Rolling and Slipping

還不明白的薑

2010-02-17 7:38 pm

A uniform solid sphere is released from rest on a rough plane inclined at 45 degree to the horizontal and the coefficient of friction is k.

(a) If the sphere rolls without slipping, show that k >= 2/7 and that the centre of the sphere has an acceleration 5\/(2) g/14 down the plane.

(b) If k<2/7 and a is the acceleration of the centre of the sphere down the plane, show that a > 4g \/(2) / 15 .

Given that moment of inertia of a uniform solid sphere of radius a about a diameter is 2ma^2 /5 .

回答 (1)

Prof. Physics

2010-02-17 8:32 pm

✔ 最佳答案

You have missed out one information, the radius of the sphere should be 1 m.

a. Normal reaction, R = mgcos45* = mg / sqrt2

Let a be the acceleration of the centre of the sphere down the plane.

mgsin45* - f = ma

mg / sqrt2 - f = ma ... (1)

Take moment about the centre

Angular acceleration = ra

Ira = fr

Ia = f

f = 2mr^2a / 5

Linear acceleration, a = 5f / 2mr^2 ... (2)

Put (2) into (1):

mg / sqrt2 = f + ma

mg / sqrt2 = f[1 + 5/2r^2] = 7f / 2

f = 2mg / 7sqrt2 ... (3)

For rolling without slipping, k >= f/R

k >= (2mg / 7sqrt2) / (mg / sqrt2)

k >= 2/7

By eqt (1): mg/sqrt2 = f + ma

mg/sqrt2 = 2ma/5 + ma

7ma / 5 = mg/sqrt2

Acceleration, a = 5gsqrt2 / 14 down the plane

b. For k < 2/7, slipping occurs before rolling.

Friction, f = kR = kmg / sqrt2

By eqt (1):

mg / sqrt2 = f + ma

mg / sqrt2 = kmg / sqrt2 + ma

g / sqrt2 < (2/7)g / sqrt2 + a

a > 5gsqrt2 / 14 > 4gsqrt2 / 15

P.S. This question is out of the current AL Applied maths syllabus

參考: Physics king

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