High School

Problem No. 2: A 5kg cart carrying a 35kg grocery is attached to a 1.5m long rope. The cart is spun in a full circle horizontally.

If the tension of the rope is 100N, how much time does it take for the cart to make one rotation?

Answer :

To find the time it takes for the cart to make one rotation, we'll need to use some basic physics concepts related to uniform circular motion.


  1. Understanding the Scenario:


    • The total mass of the cart and groceries is [tex]5 \text{ kg} + 35 \text{ kg} = 40 \text{ kg}[/tex].

    • The tension in the rope, [tex]T[/tex], acts as the centripetal force that keeps the cart moving in a circle.

    • The length of the rope is the radius [tex]r[/tex] of the circle, [tex]r = 1.5 \text{ m}[/tex].



  2. Apply the formula for centripetal force:


    • The formula for centripetal force [tex]F_c[/tex] is:
      [tex]F_c = \frac{mv^2}{r}[/tex]

    • Here, [tex]m[/tex] is the mass, [tex]v[/tex] is the velocity, and [tex]r[/tex] is the radius of the circle.

    • Set the centripetal force equal to the tension in the rope: [tex]100 \text{ N} = \frac{40 \cdot v^2}{1.5}[/tex]



  3. Solve for velocity [tex]v[/tex]:


    • Rearrange the equation to solve for [tex]v[/tex]:
      [tex]100 \times 1.5 = 40 \cdot v^2[/tex]
      [tex]150 = 40 \cdot v^2[/tex]
      [tex]v^2 = \frac{150}{40} = 3.75[/tex]
      [tex]v = \sqrt{3.75} \approx 1.936 \text{ m/s}[/tex]



  4. Find the time for one complete rotation:


    • The formula for the period [tex]T[/tex] (time for one complete rotation) is based on the circumference of the circle:

    • The circumference [tex]C[/tex] is [tex]2 \pi r = 2 \pi \times 1.5[/tex].

    • [tex]T = \frac{C}{v} = \frac{2 \pi \times 1.5}{1.936} \approx 4.876 \text{ seconds}[/tex]




Therefore, it takes approximately 4.88 seconds for the cart to make one full rotation.

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