Question 643: Centre of Mass and Rotational Motion

Question

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s$$-$$2. Its net acceleration in m s$$-$$2 at the end of 2.0 s is approximately

Accepted Answer

Given, r = 50 cm = 0.5 m, $$\alpha $$ = 2.0 rad s
Ã¢ÂÂ2
,  $${\omega _0}$$ = 0
At the end of 2 s,
Tangential acceleration, at
= r$$\alpha $$ = 0.5 ÃÂ 2 = 1 m s
Ã¢ÂÂ2
Radial acceleration, = $${a_r} = {\omega ^2}r = {\left( {{\omega _0} + \alpha t} ight)^2}r$$
= (0 + 2 ÃÂ 2)
2
ÃÂ 0.5 = 8 m s
Ã¢ÂÂ2
$$ 	herefore $$ Net acceleration,
$$a = \sqrt {a_t^2 + a_r^2}  = \sqrt {{1^2} + {8^2}}  = \sqrt {65}  \approx 8\,m{s^{ - 2}}$$

Answer

6.0

Answer

3.0

Answer

8.0

Answer

7.0

Centre of Mass and Rotational Motion

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