
A motorcyclist of mass $m$ is to negotiate a curve of radius $r$ with a speed $v$. The minimum value of the coefficient of friction so that this negotiation may take place safely is?
A. ${v^2}rg$
B. $\dfrac{{{v^2}}}{{gr}}$
C. $\dfrac{{gr}}{{{v^2}}}$
D. $\dfrac{g}{{{v^2}r}}$
Answer
146.4k+ views
Hint: Since there is a downward force (which equals the weight of the body) applied to a body. When a body takes a curve with radius $r$ there is a chance of slipping a body hence, to negotiate the slip the normal force acting opposite to the downward force must be balanced by the centrifugal force.
Complete answer:
Mass of a motorcyclist $ = m$ (given)
Since the gravity $g$ acts downward. Therefore, the weight of a body $ = mg$
Let us consider the normal force $N$ is equal to and opposite to the direction of weight$(mg)$ of a body.
Now, we know that Centrifugal Force acting on the body:
${F_c} = m\dfrac{{{v^2}}}{r}$ where,
v = speed of a body
r = radius of curve taken by body
Also, we know that $F = \mu N$where,
F = Frictional Force and$\mu $= Coefficient of friction
To avoid slip, the frictional force must be balanced by centrifugal force i.e.,
$F = {F_c}$
$\mu N = m\dfrac{{{v^2}}}{r}$
Substitute $N = mg$ in the above expression, we get
$\mu (mg) = m\dfrac{{{v^2}}}{r}$
$\mu = \dfrac{{{v^2}}}{{gr}}$
Thus, the minimum value of the coefficient of friction so that this negotiation may take place safely is $\mu = \dfrac{{{v^2}}}{{gr}}$.
Hence, the correct option is (B) $\mu = \dfrac{{{v^2}}}{{gr}}$ >
Note: Since this is a problem based on the balancing of two different forces hence, given conditions are to be analyzed very carefully and only after which the procedure of solving the problem is identified. To have a better understanding of the formulas used, it is essential to understand which kind of forces influences the problem.
Complete answer:
Mass of a motorcyclist $ = m$ (given)
Since the gravity $g$ acts downward. Therefore, the weight of a body $ = mg$
Let us consider the normal force $N$ is equal to and opposite to the direction of weight$(mg)$ of a body.
Now, we know that Centrifugal Force acting on the body:
${F_c} = m\dfrac{{{v^2}}}{r}$ where,
v = speed of a body
r = radius of curve taken by body
Also, we know that $F = \mu N$where,
F = Frictional Force and$\mu $= Coefficient of friction
To avoid slip, the frictional force must be balanced by centrifugal force i.e.,
$F = {F_c}$
$\mu N = m\dfrac{{{v^2}}}{r}$
Substitute $N = mg$ in the above expression, we get
$\mu (mg) = m\dfrac{{{v^2}}}{r}$
$\mu = \dfrac{{{v^2}}}{{gr}}$
Thus, the minimum value of the coefficient of friction so that this negotiation may take place safely is $\mu = \dfrac{{{v^2}}}{{gr}}$.
Hence, the correct option is (B) $\mu = \dfrac{{{v^2}}}{{gr}}$ >
Note: Since this is a problem based on the balancing of two different forces hence, given conditions are to be analyzed very carefully and only after which the procedure of solving the problem is identified. To have a better understanding of the formulas used, it is essential to understand which kind of forces influences the problem.
Recently Updated Pages
Product to Sum Formulae - Important Concepts and Tips for JEE

Sum to Product Formulae - Important Concepts and Tips for JEE

Percentage Composition - Important Concepts and Tips for JEE

How to find Oxidation Number - Important Concepts for JEE

Derivative Examples – Meaning, Examples and Calculus Derivatives

Hybridization – Definition, Shape, Types and Important FAQs

Trending doubts
JEE Main Marks Vs Percentile Vs Rank 2025: Calculate Percentile Using Marks

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

JEE Main Counselling 2025: Registration and Important Dates

JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Response Sheet 2025 Released – Download Links, and Check Latest Updates

JEE Main 2025 Session 1 Result Out: Final Answer Key Released at jeemain.nta.nic.in

Other Pages
Mahavir Jayanti 2025 – Date, Significance, and Celebrations

Good Friday and Easter Celebration: Understanding the Significance of the Holy Week

Dr B. R. Ambedkar Short Biography

CBSE Class 12 English Core Syllabus 2024-25: Updated Curriculum

CBSE Date Sheet 2025 Released for Class 12 Board Exams, Download PDF

Celebrate the Joy of New Beginnings with Subho Noboborsho: A Guide to Bengali New Year Traditions
