Rolling Under The Influence

A homogeneous ball with mass m = 1 g m=1 ~\mbox{g} and total charge q = 1 0 3 C q=10^{-3}~\mbox{C} (uniformly distributed) is placed on a horizontal x y xy -plane. The ball starts rolling without slipping under the influence of a uniform electrostatic field E = ( E , 0 , 0 ) \vec{E}=(E,0,0) with E = 1 V/m . E=1~ \mbox{V/m}. Find the acceleration in meters per second squared of the center of mass of the ball.


The answer is 0.714.

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1 solution

Nishant Sharma
Mar 16, 2014

free-body diagram free-body diagram

The free-body diagram looks like the one above( sorry for the poor clarity of the same). From there we can write equations of motion as:

t r a n s l a t i o n \mathbf{translation} :

m a = q E f ( i ) ma=\,qE-f\;\;\;---(i)

r o t a t i o n a b o u t c e n t e r o f m a s s : \mathbf{rotation\,about\,center\,of\,mass:}

f r = I α fr=\,I\alpha

f r = 2 m r 2 α 5 \rightarrow\,fr=\,\displaystyle\frac{2mr^2\alpha}{5}

f r = 2 m r 2 a 5 r \rightarrow\,fr=\,\displaystyle\frac{2mr^2a}{5r} ( ( Since for pure rotation/no slipping a = α r ) a=\alpha\,r)

f = 2 m a 5 ( i i ) \rightarrow\,f=\displaystyle\frac{2ma}{5}\;\;\;----(ii)

From ( i i ) (ii) and ( i ) (i) we have,

7 m a 5 = q E \displaystyle\frac{7ma}{5}=qE

a = 5 q E 7 m \rightarrow\,a=\displaystyle\frac{5qE}{7m}

Plugging in the values in S.I. we have a = 5 7 \displaystyle\,a=\boxed{\frac{5}{7}} .

yeah ! a good one .......... I did not think about the moment of inertia

Mayankk Bhagat - 7 years, 2 months ago

good visualiation

Sahil Kharb - 7 years, 1 month ago

did in same way

Priyesh Pandey - 7 years ago

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