A bob is whirled in a horizontal plane by means of a string with an initial speed of o rpm.
The tension in the string is T. If speed becomes 20 while keeping the same radius, the tension in the string becomes :
T
1) T
(2) 4T
(3)
(4) V2T
When a bob is whirled in a horizontal plane using a string, the tension in the string acts as the centripetal force required to keep the bob moving in its circular path. This tension is crucial for maintaining the bob’s circular motion.
The centripetal force needed to keep an object moving in a circle is determined by the formula:
\[ F_c = \frac{m v^2}{r} \]
where:
- \( m \) is the mass of the bob,
- \( v \) is the linear speed of the bob,
- \( r \) is the radius of the circular path.
In this scenario, if the bob is initially moving at a certain speed and the tension in the string is \( T \), then the relationship between these quantities is:
\[ T = \frac{m v^2}{r} \]
If the speed of the bob changes while keeping the radius of the circle constant, the tension in the string will also change. Specifically, the tension is proportional to the square of the speed. Therefore, if the speed of the bob increases, the tension in the string increases by the square of the factor by which the speed increased. Similarly, if the speed decreases, the tension decreases by the square of the factor by which the speed decreased.
For example, if the bob’s speed is increased to a new value, the new tension in the string can be calculated based on this squared relationship. This means that if the speed is doubled, the tension becomes four times greater; if the speed is tripled, the tension becomes nine times greater, and so on. Thus, the change in speed directly influences the change in tension in a squared relationship, reflecting the fundamental nature of centripetal force in circular motion.
To solve this problem, let's start by understanding how the tension in the string relates to the speed of the bob.
When an object is whirled in a horizontal plane, the tension in the string provides the centripetal force needed to keep the object moving in a circular path. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( m \) is the mass of the bob, \( v \) is the speed of the bob, and \( r \) is the radius of the circle.
Initially, the bob is moving with a speed of \( v_0 \) (in rpm) and the tension in the string is \( T \). Therefore, we have:
\[ T = \frac{m v_0^2}{r} \]
When the speed increases to 20 rpm, the new tension \( T' \) can be found using:
\[ T' = \frac{m (20)^2}{r} \]
Given that:
\[ T = \frac{m v_0^2}{r} \]
To find the relationship between the initial and new tensions, we need to compare:
\[ T' = \frac{m (20)^2}{r} = \frac{m v_0^2}{r} \times \frac{(20)^2}{v_0^2} = T \times \frac{400}{v_0^2} \]
Here, \( (20)^2 = 400 \). So,
\[ T' = 400 \times \frac{T}{v_0^2} \]
If \( v_0 = 10 \), then:
\[ T' = 400 \times T \]
But if the problem asks for a specific change in speed and the options are provided (such as \( T \), \( 4T \), \( 2T \), \( \sqrt{2}T \)), then the given answer would typically match one of these options based on \( v_0 \). If the increase in speed \( v_0 \) matches the new speed exactly or follows a standard answer from options, we can simplify the ratio:
For example, if the initial speed was \( 10 \) rpm, then:
\[ T' = 4T \]
Because \( (20)^2 / (10)^2 = 4 \), so:
\[ T' = 4T \]
Thus, the correct answer is:
*(2) 4T*
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