Swinging Mass: Angular Velocity & Period

[Hyperreality]

A small mass m is attached to a point O by an inextensible string of length a. The mass is held with the string taut, at the same level as O, and released. Determine the angular velocity of the mass when the string makes an angle $$\theta$$with the downwards vertical. From this relationship, determine the period of one complete swing. How does this time compare with the value of the period of a small pendulum of length a?

My solution is
[tex]mga = \frac{1}{2}m\omega^2a^2 + mga(1-\cos\theta)[\tex]<br /> <br /> $$\omega^2 = \frac{2g\cos\theta}{a}$$<br /> <br /> But the answer says <br /> <br /> $$\omega^2=\frac{2g\sin\theta}{a}$$<br /> <br /> For the second part on the period. Assuming the answer given is right.<br /> [tex]T^2=\frac{4\pi^2}{\omega^2}[\tex]<br /> <br /> [tex]T = \pi\sqrt\frac{2a}{g\sin\theta}[\tex]<br /> <br /> But instead the answer says<br /> <br /> $$T=7.04\sqrt\frac{a}{g}$$<br /> <br /> Can anyone please tell me what I've done wrong?[/tex][/tex][/tex]

 

[Doc Al]

not SHM

A small mass m is attached to a point O by an inextensible string of length a. The mass is held with the string taut, at the same level as O, and released. Determine the angular velocity of the mass when the string makes an angle $$\theta$$with the downwards vertical. From this relationship, determine the period of one complete swing. How does this time compare with the value of the period of a small pendulum of length a?

My solution is
$$mga = \frac{1}{2}m\omega^2a^2 + mga(1-\cos\theta)$$

$$\omega^2 = \frac{2g\cos\theta}{a}$$

But the answer says

$$\omega^2=\frac{2g\sin\theta}{a}$$

For the second part on the period. Assuming the answer given is right.
$$T^2=\frac{4\pi^2}{\omega^2}$$

$$T = \pi\sqrt\frac{2a}{g\sin\theta}$$

But instead the answer says

$$T=7.04\sqrt\frac{a}{g}$$

Can anyone please tell me what I've done wrong?

Assuming your definition of the angle θ, I agree with your answer for ω. The book's answer is obviously wrong: plug in θ = 0 and you'd get zero where ω should be maximum.

To find the period, it looks like you tried to apply a relationship for simple harmonic motion, interpreting ω as the angular frequency. This is not correct, and furthermore this large amplitude pendulum does not exhibit simple harmonic motion.

Instead, to find the period you need to integrate over an entire period (or just from π/2 to 0 and muliply by 4). (In any case, your answer should not have θ in it: that's your variable of integration.)

 

[M98Ranger]

Your solution is incorrect because you have used the wrong trigonometric function. The correct equation for the angular velocity is \omega^2 = \frac{2g\sin\theta}{a}, not \cos\theta. This is because the force of gravity is acting downwards, causing the mass to swing in a circular motion, which is described by the sine function.

For the second part on the period, your equation is also incorrect because you have used the wrong value for the angular velocity. The correct equation is T = 2\pi\sqrt\frac{a}{g\sin\theta}, not \omega^2. This is because the period is the time it takes for one complete swing, which is equal to the time it takes for the mass to travel a distance of 2\pi radians. Therefore, the correct equation should be T = 2\pi\sqrt\frac{a}{g\sin\theta}.

To compare the period with that of a small pendulum of length a, we can use the equation T = 2\pi\sqrt\frac{l}{g}, where l is the length of the pendulum. Since a is the length of our swinging mass, we can see that the period of the swinging mass is equal to the period of a small pendulum of length a. This makes sense because both systems are undergoing simple harmonic motion, and the period is only dependent on the length and the acceleration due to gravity.