Trig problem

**jo74** · June 28th 2009, 09:02 AM

Hi guys. I need some help to solve this problem.
a) A body of mass m kg is attached to a point by string of length
1.25 m. If the mass is rotating in a horizontal circle 0.75 m below the
point of attachment, calculate its angular velocity.

(b) If the mass rotates on a table, calculate the force on the table when
the speed of rotation is 25 rpm and the mass is 6 kg

**skeeter** · June 28th 2009, 01:02 PM

Originally Posted by jo74

Hi guys. I need some help to solve this problem.
a) A body of mass m kg is attached to a point by string of length
1.25 m. If the mass is rotating in a horizontal circle 0.75 m below the
point of attachment, calculate its angular velocity.

(b) If the mass rotates on a table, calculate the force on the table when
the speed of rotation is 25 rpm and the mass is 6 kg

part (a)

$r = \sqrt{1.25^2 - .75^2}$

$\theta = \arcsin\left(\frac{.75}{1.25}\right)$

if $T$ is the tension in the string ...

$T\sin{\theta} = mg$

$T\cos{\theta} = F_c$ , where $F_c$ is the centripetal force.

eliminate $T$ in the two equations above and solve for $F_c$ ... then use the equation below to determine $\omega$

$F_c = mr\omega^2$

part (b) is too easy ... think about the net force in the vertical direction.

**ady72** · August 11th 2009, 11:10 AM

Hi it's me again. I tried to find the answer but just got my paper back but got it completely wrong again. I already had the first part answered, it's the second part I'm struggling with.

**skeeter** · August 11th 2009, 11:32 AM

Originally Posted by jo74

Hi guys. I need some help to solve this problem.
a) A body of mass m kg is attached to a point by string of length
1.25 m. If the mass is rotating in a horizontal circle 0.75 m below the
point of attachment, calculate its angular velocity.

(b) If the mass rotates on a table, calculate the force on the table when
the speed of rotation is 25 rpm and the mass is 6 kg

is the string still attached as in part (a) ?

**ady72** · August 11th 2009, 11:40 AM

Yeah, it is still attached to the string at the same angle as in part 1.

**skeeter** · August 11th 2009, 03:09 PM

$T\cos{\theta} = m r \omega^2$

$T = \frac{m r \omega^2}{\cos{\theta}}$

sub in your known values and calculate $T$

let $F$ = force that the table exerts upward on the mass.

$T\sin{\theta} + F = mg$

$F = mg - T\sin{\theta}$

calculate $F$

**ady72** · August 11th 2009, 09:44 PM

Thanks for that one, much appreciated

**srobrien** · November 7th 2009, 06:58 AM

Hi I am also having problems with this exact question, I have the value for the radius and the angle but I am completly lost how to find the other values, all formulas contain m but it is an unknown value?

Thanks

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