Math Problem Statement

A small block with mass m slides on the frictionless track shown in the sketch, starting from rest at point A. The circular loop has radius R and the block starts at an initial height equal to three times the radius, h = 3R. When the block passes through the top part of the loop (point B), what is the normal force (magnitude and direction) from the track acting on the block?

Solution

This problem involves energy conservation and circular motion dynamics. Let's solve it step-by-step.

Given:

  • A small block of mass mm slides on a frictionless track.
  • The block starts at height h=3Rh = 3R (where RR is the radius of the loop).
  • We are asked to find the normal force at the top of the loop, point BB, where the block's height is 2R2R (since it is at the top of the loop).

Step 1: Conservation of Mechanical Energy

Since the track is frictionless, mechanical energy is conserved. The block starts from rest, so the total mechanical energy is purely gravitational potential energy at point AA (height h=3Rh = 3R).

  • At point A (starting point): Etotal=Potential Energy at A=mgh=mg(3R)E_{\text{total}} = \text{Potential Energy at } A = mgh = mg(3R)

  • At point B (top of the loop): The block will have both kinetic energy and potential energy: Etotal=Kinetic Energy at B+Potential Energy at BE_{\text{total}} = \text{Kinetic Energy at } B + \text{Potential Energy at } B Etotal=12mvB2+mg(2R)E_{\text{total}} = \frac{1}{2}mv_B^2 + mg(2R) where vBv_B is the velocity of the block at point BB.

Step 2: Apply Conservation of Energy

Setting the total energy at AA equal to the total energy at BB: mg(3R)=12mvB2+mg(2R)mg(3R) = \frac{1}{2}mv_B^2 + mg(2R) Solve for vB2v_B^2: mg(3R)mg(2R)=12mvB2mg(3R) - mg(2R) = \frac{1}{2}mv_B^2 mgR=12mvB2mgR = \frac{1}{2}mv_B^2 vB2=2gRv_B^2 = 2gR

Step 3: Forces at the Top of the Loop (Point B)

At the top of the loop, the block experiences two forces:

  1. Gravitational force mgmg (downward).
  2. Normal force NN from the track (also downward at point BB).

The net force provides the centripetal force required to keep the block moving in a circle: Fnet=N+mg=mvB2RF_{\text{net}} = N + mg = \frac{mv_B^2}{R} Substitute vB2=2gRv_B^2 = 2gR from the energy conservation step: N+mg=m(2gR)RN + mg = \frac{m(2gR)}{R} N+mg=2mgN + mg = 2mg Solve for NN: N=2mgmg=mgN = 2mg - mg = mg

Conclusion:

The normal force at point BB is mgmg and it acts downward. Thus, the correct answer is: (d) mgmg downward.

Would you like any additional details or explanations?

Related Questions:

  1. How does the normal force change at different points in the loop?
  2. What would happen if friction were introduced on the track?
  3. How would the normal force change if the block started from a lower height, say h=2Rh = 2R?
  4. Can the block complete the loop if its starting height is less than 2R2R?
  5. How does the velocity of the block change as it moves along the track?

Tip:

When solving problems involving circular motion, always account for the centripetal force, which is the net inward force keeping an object moving in a circle.

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Math Problem Analysis

Mathematical Concepts

Conservation of Mechanical Energy
Circular Motion
Forces and Dynamics

Formulas

Conservation of Energy: E_total = Potential Energy + Kinetic Energy
Potential Energy: PE = mgh
Kinetic Energy: KE = 1/2 mv^2
Centripetal Force: F = mv^2 / R

Theorems

Energy Conservation Theorem
Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12 (Physics)