A uniform ring of mass $m$ , with the outside radius $r_{2}$ , is fitted tightly on a shaft of radius $r_{1}$ . The shaft is rotated about its axis with a constant angular acceleration $β$ . Find the moment of elastic forces in the ring as a function of the distance $r$ from the rotation axis.

N = \frac{1}{2} \frac{m β}{(r_{2}^{2} - r_{1}^{2})} (r^{4} - r_{1}^{4})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

N = \frac{1}{2} \frac{m β}{(r_{2}^{2} + r_{1}^{2})} (r_{2}^{4} - r^{4})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N = \frac{1}{2} \frac{m β}{(r_{2}^{2} + r_{1}^{2})} (r_{2}^{4} + r^{4})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $N = \frac{1}{2} \frac{m β}{(r_{2}^{2} - r_{1}^{2})} (r^{4} - r_{1}^{4})$
Moment applied by the elastic force due to inside and outside part on the elementary ring of radius $r$ are $N$ and $N + \frac{d N}{d r} d r$ respectively.
$∴$ Net moment of force is $(N + \frac{d N}{d r} d r - N) = \frac{d N}{d r} d r = d N$
Given: Mass of the ring is $m$
Area of cross-section of the ring $A = π r_{2}^{2} - π r_{1}^{2} = π (r_{2}^{2} - r_{1}^{2})$
Area of the cross-section of the elementary ring $A^{'} = 2 π r d r$
$∴$ Mass of the elementary ring $d m = \frac{m}{A} A^{'} = \frac{m}{π (r_{2}^{2} - r_{1}^{2})} 2 π r d r$

Now moment of inertia of the elementary part $d I = r^{2} d m = \frac{m}{π (r_{2}^{2} - r_{1}^{2})} 2 π r^{3} d r$
Moment of force on the elementary part $d N = d I \times β$
$∴$ $d N = \frac{2 m β}{(r_{2}^{2} - r_{1}^{2})} r^{3} d r$
Total moment of force on the ring $N = \frac{2 m β}{(r_{2}^{2} - r_{1}^{2})} \int_{r_{1}}^{r} r^{3} d r$
$N = \frac{2 m β}{(r_{2}^{2} - r_{1}^{2})} \times \frac{r^{4}}{4} {∣ ∣ ∣}_{r_{1}}^{r}$

$N = \frac{m β}{2 (r_{2}^{2} - r_{1}^{2})} (r^{4} - r_{1}^{4})$

Suggest Corrections

Similar questions

R_{1}

R_{2}

R_{1} + R_{2} = R

R_{1}^{2} + R_{2}^{2} = R^{2}

R_{1} + R_{2} < R

R_{1}^{2} + R_{2}^{2} < R_{2}

μ = \frac{q}{2 m}

r

R

M

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

If the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle of radius R then

(a) $R_{1} + R_{2} = R$

(b) $R_{1} + R_{2} > R$

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

Join BYJU'S Learning Program