A 0.35 kg stone attached to a 0.8 m long string is rotated in a horizontal plane. The string makes an angle of 20° with the horizontal. Determine the speed of the stone.
I've tried tan 20= gr/v^2 = (9.8)(.8)/v^2 = 4.7 m/s and that's not the correct answer (online homework). Any help or explanation would be appreciated
mg=Tsin α
mv²/R=Tcos α
mgR/ mv² = Tsin α/Tcos α
gR/ v² =tan α
gLcos α/ v² =tan α
v=sqrt{ gLcos α/ tan α} =sqrt{9.8•0.8•cos20°/tan20°} = 4.5 m/s
To determine the speed of the stone, we can use the concept of centripetal acceleration.
Centripetal acceleration is given by the formula:
a = v^2 / r
where v is the speed of the stone and r is the radius of the circular path (in this case, the length of the string).
In the problem, the string makes an angle of 20° with the horizontal. This means the radius of the circular path can be found using trigonometry.
Using the relationship between the angle and the sides of a right triangle, we can say:
sin(20°) = opposite / hypotenuse
sin(20°) = r / 0.8 m
Rearranging the equation, we find:
r = 0.8 m * sin(20°)
r ≈ 0.28 m
Now, substituting the radius and the given value for acceleration due to gravity (g = 9.8 m/s^2) into the centripetal acceleration formula, we get:
a = v^2 / r
9.8 m/s^2 = v^2 / 0.28 m
Rearranging the equation and solving for v, we find:
v^2 = 9.8 m/s^2 * 0.28 m
v^2 ≈ 2.744 m^2/s^2
Taking the square root of both sides, we find:
v ≈ √(2.744 m^2/s^2)
v ≈ 1.656 m/s
Therefore, the speed of the stone is approximately 1.656 m/s.