Questions: A ball weighing 15.2 N is tied to a string fixed to the ceiling. The string makes a 27.9° angle with the ceiling. Initially, the ball is held in place by a force that is perpendicular to the string. θ₁=27.9° Just after the ball is released and allowed to start swinging back and forth, what is the tangential acceleration of the ball? m / s²

Solution Steps

The forces acting on the ball are:

• The gravitational force (\( F_g \)) acting downward.
• The tension in the string (\( T \)).
• The perpendicular force (\( F \)) holding the ball in place initially.

Resolve the gravitational force into components parallel and perpendicular to the string:

• The parallel component (\( F_{g,\parallel} \)) is \( F_g \sin(\theta_1) \).
• The perpendicular component (\( F_{g,\perp} \)) is \( F_g \cos(\theta_1) \).

The tangential acceleration (\( a_t \)) is caused by the parallel component of the gravitational force: \[ a_t = \frac{F_{g,\parallel}}{m} = \frac{F_g \sin(\theta_1)}{m} \]

Given:

• \( F_g = 15.2 \, \text{N} \)
• \( \theta_1 = 27.9^\circ \)
• \( g = 9.8 \, \text{m/s}^2 \)

First, find the mass (\( m \)) of the ball: \[ m = \frac{F_g}{g} = \frac{15.2 \, \text{N}}{9.8 \, \text{m/s}^2} = 1.55 \, \text{kg} \]

Now, calculate the tangential acceleration: \[ a_t = \frac{15.2 \, \text{N} \cdot \sin(27.9^\circ)}{1.55 \, \text{kg}} \]

Final Answer