the 200g head of a golf club moves 40 m/s in a circular arc of 1.2m radius. How much force must the player exert on the handle of the club to prevent it from flying out of her hands at the bottom of the swing? ignore the mass of the clubs shaft.
F=mv²/R
The 200.g head of a golf club moves at 40 m/s in a circular ARC of 1.2 m radius. How much force must the player exert on the handle of the club to prevent it from flying out of her hands at the bottom of the swing? Ignore the mass of the club's shaft.
To find the force the player must exert on the handle of the golf club, we can use the principles of centripetal force.
Centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle and its magnitude is given by the equation:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the object (200 grams, or 0.2 kg in this case)
v is the velocity of the object (40 m/s in this case)
r is the radius of the circular path (1.2 meters in this case)
First, we need to convert the mass from grams to kilograms. Since 1 kilogram is equal to 1000 grams, the mass of the object is 0.2 kg.
Next, we can substitute the given values into the equation:
F = (0.2 kg * (40 m/s)^2) / 1.2 m
F = (0.2 kg * 1600 m^2/s^2) / 1.2 m
F = 266.67 N
Therefore, the player must exert a force of approximately 266.67 Newtons on the handle of the golf club to prevent it from flying out of her hands at the bottom of the swing.