A non conducting uniform solid sphere of mass m and radius R having uniform volume charge density $+ ρ$ and $- ρ$ in two hemispherical halves as shown in figure. It is then placed on a rough non conducting horizontal plane. At t = 0 a uniform electric field $\vec{E} = E_{0} \hat{i} N / C$ is switched on and the solid sphere starts rolling without sliding. The magnitude of frictional force at t = 0 is $\frac{2 ρ π R^{3} E_{0}}{K}$ where K is a positive integer. Find the value of K. (Take $\tan 53 ° = \frac{4}{3}$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Charges can be assumed to be concentrated at center of mass of respective hemispheres.

Distance of COM from center of sphere is $d = \frac{3 R}{8}$ .

$q = d e n s i t y \times v o l u m e o f h e m i s p h e r e = ρ \frac{2}{3} π R^{3}$

Dipole Moment, $P = q 2 d = (ρ \frac{2}{3} π R^{3}) (\frac{3 R}{4})$

$\begin{array}{l} τ_{Q} = P E \sin 53 ° = (ρ \frac{2}{3} π R^{3}) (\frac{3 R}{4}) E_{0} \frac{4}{5} = \frac{2}{5} ρ π R^{4} E_{0} \\ B U T \\ T o r q u e, τ = I α = (\frac{7}{5} m R^{2} α) \\ \Rightarrow \frac{2}{5} ρ π R^{4} E_{0} = (\frac{7}{5} m R^{2} α) \\ \Rightarrow α = \frac{2}{7} \frac{ρ π R^{2} E_{0}}{m} \\ a n d \\ a_{c m} = R α = \frac{2}{7} \frac{ρ π R^{3} E_{0}}{m} \\ f = m a_{c m} = \frac{2}{7} ρ π R^{3} E_{0} \end{array}$