John Abbott College Mechanics tutoring

Find the vertical drop, then use conservation of mechanical energy. At the lowest point, the net force toward the pivot supplies the inward acceleration.

1. 1. Find the vertical drop

   Let $h$ be the release height above the lowest point, in metres. The initial vertical distance below the pivot is $L\cos\theta$; at the bottom it is $L$.

   $$h=L-L\cos\theta$$

   $$h=L(1-\cos\theta)$$

   $$h=(0.80\,\mathrm m)(1-\cos60^\circ)$$

   $$h=(0.80\,\mathrm m)(1-0.50)$$

   $$h=0.40\,\mathrm m$$

   Use the vertical height, not the distance travelled along the arc.

2. 2. Use energy to find speed

   Let $v$ be the speed at the lowest point, in $\mathrm{m/s}$. Choose zero gravitational potential energy at that point. With the mass and Earth as the energy system, the initial gravitational potential energy becomes kinetic energy. The string’s tension is perpendicular to motion and does no work.

   $$mgh=\frac12mv^2$$

   $$v^2=2gh$$

   $$v^2=2(9.8)(0.40)\,\mathrm{m^2/s^2}$$

   $$v^2=7.84\,\mathrm{m^2/s^2}$$

   Taking the positive square root gives:

   $$v=2.8\,\mathrm{m/s}$$

3. 3. Change to a force equation at the bottom

   Let $T$ be the string tension in newtons. At the lowest point, draw $T$ upward toward the pivot and $mg$ downward. The path is circular, so the inward acceleration is $a_r=\frac{v^2}{L}$; the subscript $r$ denotes the radial direction. Here that direction is upward.

   $$a_r=\frac{7.84\,\mathrm{m^2/s^2}}{0.80\,\mathrm m}$$

   $$a_r=9.8\,\mathrm{m/s^2}$$

   Even though the velocity is horizontal at this instant, its direction is changing.

4. 4. Solve for tension

   Newton’s second law toward the pivot is:

   $$T-mg=\frac{mv^2}{L}$$

   Thus:

   $$T=mg+\frac{mv^2}{L}$$

   Calculate the two terms using SI values:

   $$mg=(0.50)(9.8)\,\mathrm N$$

   $$mg=4.9\,\mathrm N$$

   $$\frac{mv^2}{L}=\frac{(0.50)(7.84)}{0.80}\,\mathrm N$$

   $$\frac{mv^2}{L}=4.9\,\mathrm N$$

   $$T=4.9\,\mathrm N+4.9\,\mathrm N$$

   $$T=9.8\,\mathrm N$$

5. 5. Explain why $T=mg$ would fail

   The weight is $4.9\,\mathrm N$, so tension is twice the weight here. If $T$ equalled $mg$, the net inward force would be zero, which would not keep this moving mass on its circular path. Tension can do zero work while still changing the direction of velocity.