Question

A metal ring of radius \(10\) cm and mass \(0.5\) kg is rolling down an inclined plane from rest without slipping. The inclined plane makes an angle of \(30^\circ\) with the horizontal. What is the linear acceleration of the ring? \[ g=10\ \text{m s}^{-2} \]

• A. \(2.5\ \text{m s}^{-2}\)
• B. \(5\ \text{m s}^{-2}\)
• C. \(10\ \text{m s}^{-2}\)
• D. \(3.33\ \text{m s}^{-2}\)

Hint:

For rolling bodies: \[ a=\frac{g\sin\theta} {1+\dfrac{I}{mR^2}} \] For a ring, \[ I=mR^2 \] which gives \[ a=\frac{g\sin\theta}{2}. \]

Answer 1

The Correct Option is A

Step 1: Rolling body acceleration formula.
A body rolling down an incline without slipping has $a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$. The extra term in the denominator accounts for the energy that goes into spinning.
Step 2: Put in the ring shape factor.
For a ring all the mass sits at the rim, so $I = mR^2$, which means $\frac{I}{mR^2} = 1$.
Step 3: Simplify the formula for a ring.
\[ a = \frac{g\sin\theta}{1+1} = \frac{g\sin\theta}{2} \]
Step 4: Note the given angle and gravity.
$\theta = 30^\circ$, so $\sin 30^\circ = \frac{1}{2}$, and $g = 10$ m per s squared.
Step 5: Substitute the values.
\[ a = \frac{10 \times \frac{1}{2}}{2} = \frac{5}{2} \]
Step 6: Final value.
$a = 2.5$ m per s squared. The mass and radius drop out, so they were not needed.
\[ \boxed{a = 2.5\ \text{m s}^{-2}} \]