A boy whirls a stone in a horizontal circle of radius 1.2 m and at height 2.0 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?
How long did it take to fall to the ground? h= 1/2 g t^2
solve for t.
Then, its velocity is 10m/time
centripetal acceleration then is ...
ok, so I did h=1/2 g t^2 and got t=.64s
Then I did V=10 / .64 = 15.7 m/s
Then I did a=v^2 / r
a = (15.7)^2 / 1.2 = 204.2 ??? that seems way too big. what am I doing wrong??
looks right to me, 204m/s^2, about 20g
ok. thanks for your help!
To find the magnitude of the centripetal acceleration of the stone while in circular motion, you correctly used the equation: a = v^2 / r, where v is the velocity of the stone and r is the radius of the circular path.
You first found the time it took for the stone to fall to the ground using the equation for height h = (1/2)gt^2, where g is the acceleration due to gravity. To solve for t, you rearranged the equation: t = √(2h / g), where h is the height of the stone above the ground level, which is 2.0 m. Plugging in the values, you correctly found t = 0.64 s.
Next, you calculated the velocity of the stone by dividing the horizontal distance it traveled (10 m) by the time it took to fall (0.64 s): v = 10 m / 0.64 s = 15.7 m/s. Until this point, your calculations are correct.
However, when you calculated the centripetal acceleration using the equation a = v^2 / r, you made an error in your calculation. Plugging in the values, it should be: a = (15.7 m/s)^2 / 1.2 m = 204.6 m/s^2 (rounded to one decimal place).
So, the correct magnitude of the centripetal acceleration of the stone while in circular motion is approximately 204.6 m/s^2.
Please note that this is a relatively large value, which means the stone experienced a significant acceleration toward the center of the circular path.