Question

A 0.18-kg ball on a stick is whirled on a vertical circle at a constant speed. When the ball is at the three o’clock position, the stick's tension is 19 N. Find the tension in the stick when the ball is (a) at the twelve o’clock and (b) at the six o’clock positions.

Answer 1

To find the tension in the stick at the twelve o’clock and six o’clock positions, consider the forces acting on the ball at each position and calculate the tension using the equation T = mg + mv^2 / r for twelve o'clock and T = mg - mv^2 / r for six o'clock.

Step-by-step explanation:

To find the tension in the stick when the ball is at the twelve o’clock and six o’clock positions, we need to consider the forces acting on the ball at each position. At the twelve o'clock position, the tension in the stick is equal to the weight of the ball plus the centripetal force required to keep it moving in a circle. The tension can be calculated using the equation T = mg + mv^2 / r, where T is the tension, m is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and r is the radius of the circle.

At the six o'clock position, the tension in the stick is equal to the weight of the ball minus the centripetal force. Using the same equation, the tension can be calculated by subtracting mv^2 / r from mg.

Substituting the given values into the equation for each position will give you the tension in the stick.

Answer 2

a) 17 N

b) 21 N

Step-by-step explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.