I have followed the sample answer for the on-line #6, but am not getting a corre

Question

I have followed the sample answer for the on-line #6, but am not getting a correct answer for parts b - d. Here is the problem: A hollow sphere of radius .160m, with rotational inertia I = .0776 kgm^2 about a line through its center of mass, rolls without slipping up a surface inclined at 39.8 degress to the horizontal. At a certain initial position, the sphere's total kinetic energy is 29 J. (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved .550 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Thanks!

Explanation / Answer

(b)for hollow sphere I = (2/3)mR2

E = 1/2mv2 + 1/2I2

= v/R

E = (1/2)mv2 + (1/2)I (v/R)2

= (1/2)(m+I/R2) v2

= (1/2)(m+2m/3)v2

= 5/6 mv2

v = sqrt(6E/5m)

= sqrt(4ER2/5I)

= 2R sqrt(E/5I)

= 2*0.16 sqrt(29/5*0.0776) = 2.767 m/s

(c) KE + mgh = KE0

KE = KE0 - mgh = KE0 - mgdsin

= KE0 -(3I/2R2)gdsin

= 29 - (3*0.0776 / 2*0.162)*0.55*sin39.8 = 27.40 J

(d) same as (b) , plug in 27.4 J for E

v = 2*0.16sqrt(27.4/5*0.0776) = 2.689 m/s

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