A small 0.469-kg object moves on a frictionless horizontal table in a circular path of radius 1.42 m. The angular speed is 6.31 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Question

A small 0.469-kg object moves on a frictionless horizontal table in a circular path of radius 1.42 m. The angular speed is 6.31 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Answer 1

Centripetal acceleration a=ω²•R=T/m

R=T/m•ω²
The law of conservation of angular momentum
I• ω²=I₀•ω₀²
ω=(I₀/I)•ω₀²=(R₀²/R²)•ω₀
R³=(m•R⁴•ω₀²)/T ,
R=...

Answer 2

what else elena...???

Answer 3

To find the radius of the smallest possible circle on which the object can move, we need to determine the tension in the string when the circle reaches its smallest radius.

First, let's calculate the centripetal force acting on the object as it moves in a circular path. The centripetal force is provided by the tension in the string.

The formula for centripetal force is:
F = m * a
where F is the centripetal force, m is the mass of the object, and a is the centripetal acceleration.

The centripetal acceleration can be determined using the formula:
a = r * ω^2
where r is the radius of the circular path, and ω is the angular velocity (angular speed) of the object.

Substituting the values given in the problem:
m = 0.469 kg,
r = 1.42 m,
ω = 6.31 rad/s

First, calculate the centripetal acceleration:
a = (1.42 m) * (6.31 rad/s)^2

Next, use the centripetal acceleration to calculate the centripetal force:
F = (0.469 kg) * [(1.42 m) * (6.31 rad/s)^2]

Now, we have to ensure that the tension in the string does not exceed 105 N. If the tension exceeds this limit, the string may break.

So, the tension in the string should be less than or equal to 105 N:
T <= 105 N

Since the centripetal force is equal to the tension in the string, we can now set up the inequality:
F <= 105 N

Finally, we can substitute the value of the centripetal force calculated earlier into the inequality and solve for the smallest possible radius:

(0.469 kg) * [(1.42 m) * (6.31 rad/s)^2] <= 105 N

Rearrange the inequality to solve for the radius:
r <= sqrt[(105 N) / (m * (6.31 rad/s)^2)]

Substituting the values:
r <= sqrt[(105 N) / (0.469 kg * (6.31 rad/s)^2)]

Evaluating this expression will give you the radius of the smallest possible circle on which the object can move.