A ball on the end of a string is whirled around
in a horizontal circle of radius 0.486 m. The
plane of the circle is 1.83 m above the ground.
The string breaks and the ball lands 2.39 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9.8 m/s^2
Find the centripetal acceleration of the ball
during its circular motion.
To find the centripetal acceleration of the ball during its circular motion, you can use the formula:
Centripetal Acceleration (a) = (v^2) / r
where v is the velocity of the ball and r is the radius of the circle.
However, we do not have the velocity directly given, so we need to find it first.
To find the velocity of the ball, we can use the concept of conservation of energy. When the string breaks, the ball is only subjected to the force of gravity, so the initial gravitational potential energy is converted into kinetic energy.
The gravitational potential energy (PE) of the ball initially is given by:
PE = m * g * h
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the plane of the circle above the ground.
The kinetic energy (KE) of the ball as it hits the ground is given by:
KE = (1/2) * m * v^2
where v is the velocity of the ball when it hits the ground.
Since energy is conserved, the initial gravitational potential energy is equal to the final kinetic energy. Therefore:
PE = KE
m * g * h = (1/2) * m * v^2
Now, we can solve for v:
v = sqrt((2 * g * h))
Substituting the given values:
v = sqrt((2 * 9.8 m/s^2 * 1.83 m))
v = 6.39 m/s
Now that we have the velocity, we can substitute it into the formula for centripetal acceleration:
a = (v^2) / r
a = (6.39 m/s)^2 / 0.486 m
a = 83.94 m/s^2
Therefore, the centripetal acceleration of the ball during its circular motion is approximately 83.94 m/s^2.
a=mv²/R=?
h=gt²/2
t=sqrt(2h/g) = sqrt(2•1.83/9.8)=0.61 s.
L=v•t = >
v=L/t=2.39/0.61=3.92 m/s
a= mv²/R= m•(3.92)²/ 0.486 =…
You need mass ‘m’