Centripetal Force

Centripetal force is an acceleration of an object moving in circular path whose direction is _________________________

• tangential to the circular path in the direction of motion

• towards the centre of the circular path

• radially outwards along the circular path

• tangential to the circular path opposite to the direction of motion

What relationship exists between the radius of a circle and the centripetal acceleration?

• $a=\frac{gm}{r\ }$

• $a=\frac{v^2}{r}$
  

• $a=\frac{2\pi v}{r^2}$
  

• $\frac{2\pi r}{T}$

Which of the following is the unit for centripetal force?

• J
• m/s²
• kg⋅m/s
• N

What is the equation for centripetal force?

•  $F=\frac{mv^2}{r}$  
  

•  $F=\frac{ma}{r^2}$  
  

•  $F=\frac{2\pi v}{T}$  
  

•  $F=\frac{2\pi r^2}{v}$  
  

A horse gallops around a circular arena.  The centripetal force on the horse is ___.

• directed outward
• tension
• directed toward the center of the arena
• 9.8 N

Tags

NGSS.HS-PS2-1

Which of the following increase the centripetal force?

• Increasing the radius
• Increasing the velocity
• Decreasing the mass
• Decreasing the acceleration

Calculate the amount of centripetal force it takes for a softball pitcher to rotate a ball with a velocity of 3 m/s if the ball weighs .12 kg and her arm is .65 meters long.

• .38
• 1.23
• .55
• 1.66 N

Tags

NGSS.HS-PS2-1

When an object is in circular motion it is constantly changing its speed.

• True

• False

What is the period of a geosynchronous satellite orbiting at a distance of 36,000 km from the Earth's surface (the radius of the Earth is r_e=6.37 x 10⁶ m).

• 86400 s

• 24 s

• No period - it stays in the same position above the Earth

• 24000 seconds

What is the velocity of a geosynchronous satellite orbiting at a distance of 36,000 km from the Earth's surface (the radius of the Earth is r_e=6.37 x 10⁶ m).

• 500 m/s

• 3100 m/s

• 11,000 m/s

• 53,000 m/s