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Gnesinka [82]
3 years ago
14

A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen

ter of the sphere is to be 0.10 gb) For this angle, what would be the acceleration of a frictionless block sliding down the incline
Physics
2 answers:
Liono4ka [1.6K]3 years ago
5 0

Answer:

Part a)

\theta = 8.05 degree

Part b)

a = 1.37 m/s^2

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

mg sin\theta - F_f = ma

also we have torque equation on it

F_f R = I\alpha

for pure rolling

a = R \alpha

F_f = \frac{Ia}{R^2}

now we have

mg sin\theta = ma + \frac{Ia}{R^2}

now we have

mg sin\theta = (m + \frac{2}{5}m)a

a = \frac{5}{7}g sin\theta

now given that

a = 0.10 g

so we have

0.10 g = \frac{5}{7} g sin\theta

sin\theta = 0.14

\theta = 8.05 degree

Part b)

If the inclined plane is frictionless then the acceleration is given as

a = g sin\theta

a = 9.8(0.14)

a = 1.37 m/s^2

velikii [3]3 years ago
5 0

Answer:

Explanation:

Given

Moment of inertia of sphere is I=\frac{2Mr^2}{5}

Here Friction will Provide torque

thus f_r\times r=I\times \alpha

f_r=\frac{I}{r^2}

Also friction will oppose weight sin component

mg\sin \theta -f_r=ma

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

a=0.1 g=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

\sin \theta =0.14

\theta =8.407^{\circ}

(b)Considering \mu the coefficient of friction

acceleration of block sliding down the incline

a=g\sin \theta -\mu g\cos \theta

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