Question 138: Work, Energy and Power & Circular Motion

Question

A body is projected at t = 0 with a velocity 10 ms
–1
at an angle of 60
o
with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms
–2
, the value of R is :

Accepted Answer

v
x
= 10cos60
o
= 5 m/s
v
y
= 10cos30
o
= $$5\sqrt 3 $$ m/s
velocity after t = 1 sec.
v
x
= 5 m/s
v
y
= $$\left| {\left( {5\sqrt 3  - 10} ight)} ight|$$ m/s = 10 $$-$$ 5$$\sqrt 3 $$
a
n
= $${{{v^2}} \over R} \Rightarrow \,R\,$$ = $${{v_x^2 + v_y^2} \over {{a_n}}}$$ = $${{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos 	heta }}$$
tan$$	heta $$  = $${{10 - 5\sqrt 3 } \over 5}$$ = 2 $$-$$ $${\sqrt 3 }$$ $$ \Rightarrow $$  $$	heta $$ = 15
o
R = $${{100\left( {2 - \sqrt 3 } ight)} \over {10\cos 15}} = 2.8m$$

Answer

2.8 m

Answer

5.1 m

Answer

2.5 m

Answer

10.3 m

Work, Energy and Power & Circular Motion

A body is projected at t = 0 with a velocity 10 ms –1 at an angle of 60 o with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms –2 , the value of R is :

Choose the correct answer: