
A body of mass 1kg tied to one end of string is revolved in a horizontal circle of radius 0.1m with a speed of 3 revolution/sec, assuming the effect of gravity is negligible, then linear velocity, acceleration and tension in the string be:
A. \[1.88\,m/s,35.5\,m/{s^2},35.5\,N\]
B. \[2.88\,m/s,45.5\,m/{s^2},45.5\,N\]
C. \[3.88\,m/s,55.5\,m/{s^2},55.5\,N\]
D. None
Answer
161.1k+ views
Hint:As the given body is revolving so we use angular terms here. The force by which the body is revolving in a circular path is determined as the centripetal force and the acceleration is known as the centripetal acceleration. To find this problem we use the relation between the terms like velocity, acceleration, radius and time.
Formula used:
Linear velocity is given as,
\[v = \dfrac{{\rm{s}}}{t}\]
Where s is the distance and t is time taken.
Centripetal acceleration is given as,
\[{a_c} = \dfrac{{{v^2}}}{r}\]
Where v is the speed or velocity of an object and r is the radius.
Tension (or centripetal force) in the string is given as,
\[T = {F_c} = \dfrac{{m{v^2}}}{r}\]
Where m is the mass of an object, v is the speed or velocity of an object and r is the radius.
Complete step by step solution:
Given mass m= 1kg
Radius, r=0.1m
Let \[N = 3\,{\rm{ }}revolution/\sec \]
Distance covered,
\[s = 3 \times 2\pi r \times 1\]
Thus, linear velocity,
\[v = \dfrac{s}{t} = 6\pi \times 0.1 = 1.88\,m/s\]
Now acceleration (or centripetal acceleration),
\[{a_c} = \dfrac{{{v^2}}}{r} \\
\Rightarrow {a_c} = \dfrac{{{{(1.88)}^2}}}{{0.1}} \\
\Rightarrow {a_c} = 35.5\,{\rm{ m/}}{{\rm{s}}^2}\]
Now tension in the string,
\[T = {F_c} \\
\Rightarrow T = \dfrac{{m{v^2}}}{r} \\
\Rightarrow T = \dfrac{{1 \times {{(1.88)}^2}}}{{0.1}} \\
\therefore T = 35.5\,N\]
Therefore, linear velocity, acceleration and tension in the string are \[1.88\,m/s,35.5\,m/{s^2},35.5\,N\].
Hence option A is the correct answer.
Note: An object must be subjected to a force in order to move, and the force reacts to an object differently depending on the sort of motion it shows. The force applied to an object in curvilinear motion and pointed at the axis of rotation or the centre of curvature is known as a centripetal force. Newtons are used to measure centripetal force.
Formula used:
Linear velocity is given as,
\[v = \dfrac{{\rm{s}}}{t}\]
Where s is the distance and t is time taken.
Centripetal acceleration is given as,
\[{a_c} = \dfrac{{{v^2}}}{r}\]
Where v is the speed or velocity of an object and r is the radius.
Tension (or centripetal force) in the string is given as,
\[T = {F_c} = \dfrac{{m{v^2}}}{r}\]
Where m is the mass of an object, v is the speed or velocity of an object and r is the radius.
Complete step by step solution:
Given mass m= 1kg
Radius, r=0.1m
Let \[N = 3\,{\rm{ }}revolution/\sec \]
Distance covered,
\[s = 3 \times 2\pi r \times 1\]
Thus, linear velocity,
\[v = \dfrac{s}{t} = 6\pi \times 0.1 = 1.88\,m/s\]
Now acceleration (or centripetal acceleration),
\[{a_c} = \dfrac{{{v^2}}}{r} \\
\Rightarrow {a_c} = \dfrac{{{{(1.88)}^2}}}{{0.1}} \\
\Rightarrow {a_c} = 35.5\,{\rm{ m/}}{{\rm{s}}^2}\]
Now tension in the string,
\[T = {F_c} \\
\Rightarrow T = \dfrac{{m{v^2}}}{r} \\
\Rightarrow T = \dfrac{{1 \times {{(1.88)}^2}}}{{0.1}} \\
\therefore T = 35.5\,N\]
Therefore, linear velocity, acceleration and tension in the string are \[1.88\,m/s,35.5\,m/{s^2},35.5\,N\].
Hence option A is the correct answer.
Note: An object must be subjected to a force in order to move, and the force reacts to an object differently depending on the sort of motion it shows. The force applied to an object in curvilinear motion and pointed at the axis of rotation or the centre of curvature is known as a centripetal force. Newtons are used to measure centripetal force.
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