Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh$-$1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s2)
Statement II : If the road is banked at an angle of 45$^\circ$, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh$-$1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.
Solution
On a horizontal ground,<br><br>${v_{\max }} = \sqrt {\mu Rg} = \sqrt {0.2 \times 2 \times 9.8} = 1.97$ m/s<br><br>= $1.97 \times {{18} \over 5}$<br><br>= 7.12 km/hr = 7.2 km/hr<br><br>Statement - 2<br><br>$${v_{\max }} = \sqrt {gr\left( {{{\tan \theta + \mu } \over {1 - \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{12} \over {0.8}}} = 19.5$$ km/hr<br><br>$${v_{\min }} = \sqrt {rg\left( {{{\tan \theta - \mu } \over {1 + \mu \tan \theta }}} \right)} = \sqrt {2 \times 9.8 \times {{0.8} \over {1.2}}} = 12.01$$ km/hr
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.