A seesaw has length 10.0 m and uniform mass 10.0 kg and is resting at an angle of 30°30° with respect to the ground (see the following figure). The pivot is located at 6.0 m. What magnitude of force needs to be applied perpendicular to the seesaw at the raised end so as to allow the seesaw to barely start to rotate?

A small 𝑚𝑏 = 09.0 ∗ 10−3 bug stands at one end of a thin uniform bar that is initially at rest on asmooth horizontal table. The other end of the bar pivots about a nail driven into the table and canrotate freely, without friction. The bar has a mass of 𝑀𝑏𝑎𝑟 = 49.0 ∗ 10−3𝑘𝑔 and is 𝐿𝑏𝑎𝑟 =100. 𝑐𝑚 in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with aspeed of 𝑣𝑏 = 21.0 𝑐𝑚𝑠 relative to the table. What is the angular speed of the bar just after the insectleaps? What is the total kinetic energy of the system just after the bug leaps? Where does thisenergy come from? (3 – sig figs)Problem 2The angular position 𝜃 of a 0.36m diameter flywheel is given by 𝜃 = (2.0 𝑟𝑎𝑑𝑠3 ) 𝑡3 . Find the 𝜃in radians and in degrees, at 𝑡1 = 2.0𝑠 and 𝑡2 = 5.0𝑠. (3 sig. figs)

A force of 100 N is used to generate a turning force. The force acts at a perpendicular distance of 2 m from the pivot. Give the symbol equation you should use to calculate the moment of this force.

A boy of weight 600 N sits on the see-saw as shown at a distance of 1.5 m from the pivot. Another boy with a weight of 400 N sits at the other end to balance the see-saw. What distance should the 400 N boy sit so that the see-saw is perfectly balanced?

A force of 2000 N is used to generate a turning force. The force acts at a perpendicular distance of 12 m from the pivot. Calculate the moment of this force.

To solve this problem, we need to determine the angular velocity of the rod at the instant \(\theta = 60^\circ\). We will use the principles of work and energy to solve this. ### Given: - Length of the rod, \( L = 2 \, \text{m} \) - Mass of the rod, \( m = 30 \, \text{kg} \) - Force applied, \( F = 40 \, \text{N} \) - Initial angle, \( \theta = 0^\circ \) - Final angle, \( \theta = 60^\circ \) - Neglect friction and the masses of blocks A and B. ### Steps to solve: 1. **Determine the work done by the force \( F \):** The work done by the force \( F \) as the rod moves from \(\theta = 0^\circ\) to \(\theta = 60^\circ\) can be calculated by considering the horizontal displacement of point B. When \(\theta = 0^\circ\), point B is at the origin. When \(\theta = 60^\circ\), the horizontal displacement \( x_B \) of point B is: \[ x_B = L \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \, \text{m} \] The work done by the force \( F \) is: \[ W = F \cdot x_B = 40 \, \text{N} \times \sqrt{3} \, \text{m} = 40\sqrt{3} \, \text{J} \] 2. **Determine the change in potential energy:** The center of mass of the rod moves vertically as the rod rotates. Initially, when \(\theta = 0^\circ\), the center of mass is at a height of \( L/2 \) from the bottom. When \(\theta = 60^\circ\), the vertical height \( h \) of the center of mass is: \[ h = \frac{L}{2} \cos(60^\circ) = \frac{2}{2} \times \frac{1}{2} = \frac{1}{2} \, \text{m} \] The change in potential energy \( \Delta PE \) is: \[ \Delta PE = m g \Delta h = 30 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times \left( \frac{1}{2} - 1 \right) \, \text{m} = 30 \times 9.81 \times \left( -\frac{1}{2} \right) = -147.15 \, \text{J} \] 3. **Apply the work-energy principle:** The work done by the force \( F \) is equal to the change in kinetic energy plus the change in potential energy: \[ W = \Delta KE + \Delta PE \] Since the rod starts from rest, the initial kinetic energy is zero. The final kinetic energy is: \[ \Delta KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia of the rod about the center of mass, and \( \omega \) is the angular velocity. For a rod rotating about its center, the moment of inertia \( I \) is: \[ I = \frac{1}{12} m L^2 = \frac{1}{12} \times 30 \, \text{kg} \times (2 \, \text{m})^2 = 10 \, \text{kg} \cdot \text{m}^2 \] Therefore: \[ 40\sqrt{3} = \frac{1}{2} \times 10 \times \omega^2 - 147.15 \] Solving for \(\omega\): \[ 40\sqrt{3} + 147.15 = 5 \omega^2 \] \[ 69.28 + 147.15 = 5 \omega^2 \] \[ 216.43 = 5 \omega^2 \] \[ \omega^2 = \frac{216.43}{5} = 43.286 \] \[ \omega = \