Question

The balance wheel of a watch oscillates with angular amplitude
1.0π rad and period 0.33 s. Find **(a)** the maximum
angular speed of the wheel, **(b)** the angular speed
of the wheel at displacement 1.0π/2 rad, and **(c)**
the magnitude of the angular acceleration at displacement 1.0π/4
rad.

Answer #1

**ans)**

**from above data that**

**we know that**

**x = A*cos wt**

**1.**

**V max = A*w**

**w = 2*pi/T**

**V max = 2*pi*A/T**

**V max = 2*pi*1.0*pi/0.33 =59.75
rad/sec**

**B.**

**now calculate the**

**x = A*cos wt**

**at given x**

**1.0*pi/2 = 1.0*pi*cos wt**

**cos wt = 1/2**

**wt = pi/3**

**So,**

**V = A*w*sin wt**

**V = 59.75*sin 60 deg**

**V = 51.745 rad/sec**

**C.**

**at given x,**

**1.0*pi/4 = 1.0*pi*cos wt**

**cos wt = 1/4,**

**Now acceleration is given by:**

**a = -A*w^2*cos wt**

**a = -1.0*pi*(2*pi/0.33)^2*(1/4)**

**a = -284.29 rad/sec^2**

**magnitude of angular acceleration is asked, So a =
+284.29 rad/sec^2**

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