7. 0/0.41 points | Previous Answers Tipler6 9.P.094 My Notes Ask Your Teacher A

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7. 0/0.41 points | Previous Answers Tipler6 9.P.094 My Notes Ask Your Teacher A hollow, thin-walled cylinder and a solid sphere start from rest and roll without slipping down an inclined plane of length 2.0 m. The cylinder arrives at the bottom of the plane 2.8 s after the sphere. Determine the angle between the inclined plane and the horizontal Because the cylinder and sphere descend from the same height, their kinetic energies at the bottom of the incline must be equal. Draw an extended free-body diagram showing the three forces acting on either the cylinder or sphere. Use Newton's second law to relate the accelerations to the angle of the incline and use a constant acceleration to relate the accelerations to the distances traveled down the incline eBook Submit Answer Save Progress Practice Another Version

Explanation / Answer

For sphere,

ma = (mg sin ?) - f ------------>[1]

where f is the frictional force.

a = a/r

f*r = I a = (2/5) mr^2* a = (2/5) mr *a

f = (2/5) ma

using this in equation 1

ma = mg sin ? - (2/5) ma

[7/5] ma = mg sin ?

a1 = [5/7] g sin ? -------------------------------------->[2]

Similarly for cylinder,

a2 = [1/2] g sin ? --------------------------------------->[3]

Using s = 1/2 a t2

a1t12 = a2t22 = 2*2 = 4 -------------------------->[4]

t22 = [a1/a2] t12 = [10/7] t12

But t2 = [t1+2.8]

Hence

[t1 +2.8]2 = t12 + 5.6t1 + 7.84 = [10/7]t12 = 1.43t12

-0.43t12  + 5.6t1 + 7.84 = 0

t1 = 14.3 s

from [4],

a1*t12 = 4

[5/7] g sin ?*14.32 = 4

sin ? = 14 / [9.8*6.182]

? = 0.16o

Angle between the inclined plane and the horizontal is 0.16o

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