Answer :
To solve this problem, we need to calculate the centripetal acceleration of the stone. We'll dissect the problem step-by-step:
1. Given Data:
- Length of the string ([tex]\( r \)[/tex]): [tex]\(80 \, \text{cm} = 0.8 \, \text{m}\)[/tex]
- Number of revolutions ([tex]\( N \)[/tex]): [tex]\(14\)[/tex] revolutions
- Time ([tex]\( t \)[/tex]): [tex]\(25 \, \text{seconds}\)[/tex]
2. Calculate the Period of One Revolution:
The period ([tex]\( T \)[/tex]) is the time it takes to complete one revolution. It is given by:
[tex]\[ T = \frac{t}{N} = \frac{25 \, \text{seconds}}{14} \approx 1.7857 \, \text{seconds} \][/tex]
3. Calculate the Frequency:
Frequency ([tex]\( f \)[/tex]) is the number of revolutions per second:
[tex]\[ f = \frac{N}{t} = \frac{14}{25} = 0.56 \, \text{revolutions/second} \][/tex]
4. Calculate the Angular Velocity:
Angular velocity ([tex]\( \omega \)[/tex]) in radians per second can be found using the formula:
[tex]\[ \omega = 2 \pi f \][/tex]
Substituting the frequency:
[tex]\[ \omega = 2 \pi \times 0.56 \approx 3.5186 \, \text{radians/second} \][/tex]
5. Calculate the Magnitude of Centripetal Acceleration:
The centripetal acceleration ([tex]\( a \)[/tex]) can be calculated using the formula:
[tex]\[ a = \omega^2 \times r \][/tex]
Substituting the known values:
[tex]\[ a = (3.5186)^2 \times 0.8 \, \text{m} \][/tex]
Simplifying the calculation:
[tex]\[ a \approx 12.378 \times 0.8 \approx 9.904 \, \text{m/s}^2 \][/tex]
Since [tex]\(1 \, \text{m/s}^2 = 100 \, \text{cm/s}^2\)[/tex], we convert the acceleration to cm/s²:
[tex]\[ a \approx 9.904 \, \text{m/s}^2 \times 100 \approx 990.4 \, \text{cm/s}^2 \][/tex]
Thus, the magnitude of the acceleration of the stone is approximately [tex]\( 990 \, \text{cm/s}^2 \)[/tex], which matches option D in the question.
1. Given Data:
- Length of the string ([tex]\( r \)[/tex]): [tex]\(80 \, \text{cm} = 0.8 \, \text{m}\)[/tex]
- Number of revolutions ([tex]\( N \)[/tex]): [tex]\(14\)[/tex] revolutions
- Time ([tex]\( t \)[/tex]): [tex]\(25 \, \text{seconds}\)[/tex]
2. Calculate the Period of One Revolution:
The period ([tex]\( T \)[/tex]) is the time it takes to complete one revolution. It is given by:
[tex]\[ T = \frac{t}{N} = \frac{25 \, \text{seconds}}{14} \approx 1.7857 \, \text{seconds} \][/tex]
3. Calculate the Frequency:
Frequency ([tex]\( f \)[/tex]) is the number of revolutions per second:
[tex]\[ f = \frac{N}{t} = \frac{14}{25} = 0.56 \, \text{revolutions/second} \][/tex]
4. Calculate the Angular Velocity:
Angular velocity ([tex]\( \omega \)[/tex]) in radians per second can be found using the formula:
[tex]\[ \omega = 2 \pi f \][/tex]
Substituting the frequency:
[tex]\[ \omega = 2 \pi \times 0.56 \approx 3.5186 \, \text{radians/second} \][/tex]
5. Calculate the Magnitude of Centripetal Acceleration:
The centripetal acceleration ([tex]\( a \)[/tex]) can be calculated using the formula:
[tex]\[ a = \omega^2 \times r \][/tex]
Substituting the known values:
[tex]\[ a = (3.5186)^2 \times 0.8 \, \text{m} \][/tex]
Simplifying the calculation:
[tex]\[ a \approx 12.378 \times 0.8 \approx 9.904 \, \text{m/s}^2 \][/tex]
Since [tex]\(1 \, \text{m/s}^2 = 100 \, \text{cm/s}^2\)[/tex], we convert the acceleration to cm/s²:
[tex]\[ a \approx 9.904 \, \text{m/s}^2 \times 100 \approx 990.4 \, \text{cm/s}^2 \][/tex]
Thus, the magnitude of the acceleration of the stone is approximately [tex]\( 990 \, \text{cm/s}^2 \)[/tex], which matches option D in the question.
