Question

A 5.73 kg ball is attached to the top of a vertical pole with a 2.01 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.63 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g 9.81 m/s?. angle: What is the tension of the string? tension N

Answer 1

Tcoso Тя Tsina Img

Given:

Length of the string

1 = 2.01m

Mass of the ball

m = 5.73kg

Speed of the ball

v = 4.63m/s

Solution:

Let tension in the string $=T$

Angle made by the string with the pole $=\theta$

Let is the radius of the circle in which the ball rotates.

r=1 sine

From the free body diagram of the ball it is clear that

T cost = mg mg cost = 1 9 Eqn(i)

The horizontal component of the force will give it the necessary centripetal force

Tsino = mv2 Tsine Isine mv2 Tsin’e = T(1 – cos?o) = my

Putting the value of $cos\theta$ from Eqn (I)

$T(1-\frac{m^2g^2}{T^2})=\frac{mv^2}{l} \\T^2-61T-3160=0 \\T=94.5N$

Putting the value of T in Eqn (I)

$\theta =53.5^{\circ}$

Answer:

Angle 0 = 53.59