Question

Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is

• A. 4
• B. 1
• C. 2
• D. 3

Hint:

When the centripetal forces of the two stones are equal, by substituting the value in the centripetal relation, the value of n can be calculated.

Answer 1

The Correct Option is C

To solve this problem, we must understand the relationship between centripetal force, mass, radius, and tangential speed in circular motion. The centripetal force \( F_c \) required to keep an object moving in a circle is given by the formula:

F_c = \frac{mv^2}{r}

where:

• m is the mass of the object,
• v is the tangential speed of the object,
• r is the radius of the circle.

Given:

• Mass of heavier stone = 2m, radius = \frac{r}{2}
• Mass of lighter stone = m, radius = r
• Let the tangential speed of the heavier stone be v_h and that of the lighter stone be v_l
• The tangential speed of the lighter stone is assumed to be n times that of the heavier stone: v_l = n \times v_h
• The centripetal forces experienced by both stones are the same.

Set up the expression for the centripetal forces:

For the heavier stone: F_{c_{h}} = \frac{2m \cdot v_h^2}{r/2} = \frac{4m \cdot v_h^2}{r}

For the lighter stone: F_{c_{l}} = \frac{m \cdot v_l^2}{r}

Equating the centripetal forces:

\frac{4m \cdot v_h^2}{r} = \frac{m \cdot v_l^2}{r}

Simplifying the equation by canceling out common terms:

4v_h^2 = v_l^2

Since v_l = n \times v_h, substitute in the equation:

4v_h^2 = (n \cdot v_h)^2

Solving for n:

4v_h^2 = n^2 \cdot v_h^2

Dividing both sides by v_h^2 (assuming v_h \neq 0):

4 = n^2

Taking the square root of both sides:

n = 2

Therefore, the value of n is 2. So, the correct answer is 2.