Like a Rolling Stone!

Electricity and Magnetism Level 3

The figure shows a non-conducting, non-uniformly charged ring with linear charge densities λ \lambda and − λ - \lambda , mass m m and radius R R . This ring is placed on a rough horizontal surface and a horizontal electric field E ⃗ \vec{E} is switched on.

If at some instant, the ring is in the position shown above and is rolling without slipping, find the minimum coefficient of friction μ m i n \mu_{min} required.

Details & Assumptions

  1. A uniform gravitational field ∣ g ⃗ ∣ = 10 m s − 2 |\vec{g}| = 10 ms^{-2} acts downward.
  2. λ = π 100 C m − 1 \lambda = \dfrac{\pi}{100} Cm^{-1}
  3. m = 1.5 k g m = 1.5 kg
  4. R = 0.15 m R = 0.15 m
  5. ∣ E ⃗ ∣ = 1000 N C − 1 |\vec{E}| = 1000 NC^{-1}


The answer is 0.314.

View wiki
Raj Magesh by Raj Magesh , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Gabriel Chacón
Jan 30, 2019

Calculate the torque of the electric force with respect to O O , the point of contact between the ring and the floor. Notice that no torque is exerted by the weight, the normal force or the friction force with respect to this point.

+ d q = λ R d θ ; d ⊥ = R ( 1 + cos ⁡ θ ) ; d F + = + d q ⋅ E ⟹ d τ = d F + ⋅ d ⊥ = λ R 2 E ( 1 + cos ⁡ θ ) d θ +dq=\lambda R d\theta; \quad d_\perp =R(1+\cos{\theta}); \quad dF_+=+dq\cdot E \implies d\tau=dF_+\cdot d_\perp=\lambda R^2E(1+\cos{\theta}) \,d\theta

τ + = ∫ 0 π 2 λ R 2 E ( 1 + cos ⁡ θ ) d θ = λ R 2 E ( π 2 + 1 ) \tau_+=\displaystyle \int_{0}^{\frac{\pi}{2}}\lambda R^2E(1+\cos{\theta})\, d\theta=\textstyle \lambda R^2E(\frac{\pi}{2}+1)

− d q = − λ R d θ ; d ⊥ = R ( 1 − cos ⁡ θ ) ; d F − = − d q ⋅ E ⟹ d τ = d F − ⋅ d ⊥ = − λ R 2 E ( 1 − cos ⁡ θ ) d θ -dq=-\lambda R d\theta; \quad d_\perp =R(1-\cos{\theta}); \quad dF_-=-dq\cdot E \implies d\tau=dF_-\cdot d_\perp=-\lambda R^2E(1-\cos{\theta}) \,d\theta

τ − = − ∫ 0 π 2 λ R 2 E ( 1 − cos ⁡ θ ) d θ = − λ R 2 E ( π 2 − 1 ) \tau_-=-\displaystyle \int_{0}^{\frac{\pi}{2}}\lambda R^2E(1-\cos{\theta})\, d\theta=-\textstyle \lambda R^2E(\frac{\pi}{2}-1)

Apply Newton's Second Law to the rotation:

τ + + τ − = I O ⋅ α , \tau_++\tau_-=I_O \cdot \alpha, \quad where I O I_O is the ring's moment of inertia with respect to O O : I O = m R 2 + m R 2 I_O=mR^2+mR^2 .

2 λ R 2 E = 2 m R 2 ⋅ α ⟹ α = λ E m ⟹ a = α R = π m s 2 2\lambda R^2E=2mR^2 \cdot \alpha \implies \alpha=\dfrac{\lambda E}{m} \implies a=\alpha R=\pi \frac{\text{m}}{\text{s}^2}

Apply Newton's Second Law to the translation (notice that the electric forces cancel out in this case):

μ m i n m g = m a ⟹ μ m i n = π g ≈ 0.314 \mu_{min} mg=ma \implies \boxed{\mu_{min}=\dfrac{\pi}{g}\approx0.314}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The ring is rotating about it's centre, so how come you used O as the axis of rotation for which a= R*alpha.

Ranjan Bhat - 1 year, 2 months ago

Log in to reply

For an object that is rotating without slipping, the bottom point of the object is momentarily stationary and can be modeled as the point about which the rest of the object is rotating. The problem can be solved with the center as the point of rotation, but the math will need to be adjusted to account for the differences in the torques.

Milan Patel - 4 months ago

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...