# Trig problem

• Jun 28th 2009, 09:02 AM
jo74
Trig problem
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
• Jun 28th 2009, 01:02 PM
skeeter
Quote:

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)

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

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

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

$\displaystyle T\sin{\theta} = mg$

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

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

$\displaystyle F_c = mr\omega^2$

part (b) is too easy ... think about the net force in the vertical direction.
• Aug 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.
• Aug 11th 2009, 11:32 AM
skeeter
Quote:

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) ?
• Aug 11th 2009, 11:40 AM
Yeah, it is still attached to the string at the same angle as in part 1.
• Aug 11th 2009, 03:09 PM
skeeter
$\displaystyle T\cos{\theta} = m r \omega^2$

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

sub in your known values and calculate $\displaystyle T$

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

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

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

calculate $\displaystyle F$
• Aug 11th 2009, 09:44 PM