Question

A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of radius 1.50 m at an angular speed of 6.00 rad/s. As the superhero swings the hammer, she then ascends vertically at a constant 2.00 m/s. What is the speed of the end of the hammer relative to the ground? What is the magnitude of acceleration of the end of the hammer? What is the direction of acceleration of the end of the hammer measured as an angle between a⃗and the direction toward the center of the circular path?

Answer 1

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

$\omega$ = Angular velocity = 6 rad/s

The horizontal component of speed is given by

$v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s$

The vertical component of speed is given by

$v_v=2\ m/s$

The resultant of the two components will give us the velocity of hammer with respect to the ground

$v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s$

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

$a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2$

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.