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CodeVID-P11-04-FRI-01
Friction & Circular Dynamics — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Kinetic friction is given by:
- A.$f_k=\mu_k N$
- B.$f_k\le\mu_k N$
- C.$f_k=\frac{N}{\mu_k}$
- D.$f_k=\mu_k N^2$
2.
Generally the coefficients satisfy:
- A.$\mu_k>\mu_s$
- B.$\mu_k=\mu_s$
- C.$\mu_k<\mu_s$
- D.$\mu_k=0$
3.
On an incline of angle $\theta$, the normal reaction is:
- A.$mg$
- B.$mg\sin\theta$
- C.$mg\cos\theta$
- D.$mg\tan\theta$
4.
The centripetal force needed for circular motion is:
- A.$\frac{mv}{r}$
- B.$\frac{mv^2}{r}$
- C.$\frac{mr}{v^2}$
- D.$mvr$
5.
The ideal banking angle satisfies:
- A.$\tan\theta=\frac{v^2}{rg}$
- B.$\tan\theta=\frac{rg}{v^2}$
- C.$\cos\theta=\frac{v^2}{rg}$
- D.$\tan\theta=\frac{v}{rg}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Why is it easier to keep an object moving than to start it from rest?
7.
A 4 kg block on a floor has $\mu_k=0.25$. Find the kinetic friction. ($g=10\ \text{m/s}^2$)
8.
Define the angle of repose.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A car of mass 1000 kg rounds a curve of radius 40 m at 8 m/s. Find the centripetal force.
10.
Find the banking angle for $v=20\ \text{m/s}$, $r=80\ \text{m}$. ($g=10\ \text{m/s}^2$)
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the expression $\tan\theta=\frac{v^2}{rg}$ for the ideal banking of a road, and use it to find the banking angle for a curve of radius 100 m designed for 30 m/s. ($g=10\ \text{m/s}^2$)
Answer Key
Section A — Multiple Choice Questions
- (A) $f_k=\mu_k N$
- (C) $\mu_k<\mu_s$
- (C) $mg\cos\theta$
- (B) $\frac{mv^2}{r}$
- (A) $\tan\theta=\frac{v^2}{rg}$
Section B — Short Answer (2 marks)
- Because kinetic friction is less than the limiting static friction ($\mu_k<\mu_s$).
- $f_k=\mu_k mg=0.25\times40=10\ \text{N}$.
- The incline angle at which a body just begins to slide; $\tan\theta=\mu_s$.
Section C — Short Answer (3 marks)
- $F=\frac{mv^2}{r}=\frac{1000\times64}{40}=1600\ \text{N}$.
- $\tan\theta=\frac{400}{800}=0.5$, so $\theta=\tan^{-1}(0.5)\approx26.6^{\circ}$.
Section D — Long Answer (5 marks)
- Resolving the normal reaction: $N\sin\theta=\frac{mv^2}{r}$ and $N\cos\theta=mg$; dividing gives $\tan\theta=\frac{v^2}{rg}=\frac{900}{1000}=0.9$, so $\theta\approx42^{\circ}$.
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