University

Physics 1

Physics 1

Fall 2024

The figure below shows a ramp and a ball that rolls along the ramp.

Sketch the (approximate) direction of the ball’s velocity and acceleration vectors as it passes the labeled points.

As the cart shown rolls toward the right, it launches a ball straight upward (relative to the cart). State whether the ball will land back on the cart, or the right or left of the cart, in the following cases.

- The carts moves at a constant velocity.
- The cart is speeding up.
- The cart is slowing.

- on the cart
- left of the cart
- right of the cart

The motion diagram shows a dot at the location of an object every second. Draw the velocity and acceleration vectors at each point.

Zack is driving past his house. He wants to toss a book out of the window and have it land in his driveway. If he lets go of the book exactly as he passes the end of the driveway, in what direction (relative to himself)) should he throw the book?

The runner has a velocity of \(5\) m/s to the right. Balls 1 and 2 are thrown toward her by friends standing on the ground. According to the runner, both balls are approaching her at \(10\) m/s. At what speeds were the balls thrown?

Ball 1 was thrown with speed \(15\)
m/s.

Ball 2 was thrown with speed \(5\)
m/s.

The figure shows three points on a rotating wheel.

- Draw the velocity vectors at each point.
- Rank in order, from largest to smallest, the angular velocities \(\omega_1\), \(\omega_2\), and \(\omega_3\).
- Rank in order, from largest to smallest, the speeds \(v_1\), \(v_2\), and \(v_3\).

\(\omega_1 = \omega_2 = \omega_3\)

\(v_3 > v_1 = v_2\)

A particle in uniform circular motion has acceleration \(a=8\) m/s². What is \(a\) if…

- The radius is doubled without changing the angular velocity?
- The radius is doubled without changing the particle’s speed?
- The angular velocity is doubled without changing the circle’s radius?

Using \(a= v^2 / r =\omega^2r\).

- \(a=16\) m/s²
- \(a=4\) m/s²
- \(a=32\) m/s²

The following figures show a rotating wheel. Determine the signs (\(+\) or \(-\)) of \(\omega\) and \(\alpha\).

A. \(\omega\) is \(+\), \(\alpha\) is \(+\)

B. \(\omega\) is \(-\), \(\alpha\) is \(+\)

C. \(\omega\) is \(+\), \(\alpha\) is \(-\)

D. \(\omega\) is \(-\), \(\alpha\) is \(-\)

The figures show the radial acceleration vector \(\vec{a_r}\) at four successive points on the trajectory of a particle moving in a counterclockwise circle.

For each, draw the tangential acceleration vector \(\vec{a_t}\) at points 2 and 3, or, if appropriate, write \(\vec{a_t }= \vec{0}\).

Also, give the sign of the angular acceleration \(\alpha\) (\(+\), \(-\) or \(0\)).

The velocity of a particle in reference frame A is \((2.0 \hat{\imath} + 3.0 \hat{\jmath})\) m/s. The velocity of reference frame B with respect to reference frame A is \(4.0 \hat{k}\) m/s, and the velocity of reference frame C with respect to B is \(2.0 \hat{\jmath}\) m/s. What is the velocity of the particle in reference frame C?

We are given

- \(\vec{v}_{PA} = (2.0 \hat{\imath} + 3.0 \hat{\jmath})\) m/s
- \(\vec{v}_{BA} = 4.0 \hat{k}\) m/s
- \(\vec{v}_{CB} = 2.0 \hat{\jmath}\) m/s

Using the relative velocity equations: \(\vec{v}_{PC} = \vec{v}_{PA} + \vec{v}_{AC}\) and \(\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}\).

Putting these together: \(\vec{v}_{PC} = \vec{v}_{PA} + \vec{v}_{AB} + \vec{v}_{BC} = \vec{v}_{PA} - \vec{v}_{BA} - \vec{v}_{CB}\)

So:

\(\vec{v}_{PC} = (2.0 \hat{\imath} + 3.0 \hat{\jmath}) - 4.0 \hat{k} - 2.0 \hat{\jmath} = (2.0 \hat{\imath} + 1.0 \hat{\jmath} - 4.0 \hat{k})\) m/s

A propeller blade at rest starts to rotate with a tangential acceleration of the tip of the blade at 3.00 m/s². The tip of the blade is 1.5 m from the axis of rotation. At \(t=0.8\) s, what is the magnitude of the total acceleration of the tip of the blade?

\(v = a_T t = (3.00)(0.8s) = 2.40\) m/s

\(a_c = {(2.4)^2}/{1.5} = 3.84\) m/s²

total acceleration magnitude: \(a = \sqrt{a_T^2 + a_c^2 }\) = 4.87 m/s²

A balloon is released from rest and flies around the room with the following acceleration:

\(\vec{a}(t) = (b-ct^2) \hat{\imath }+ (q + nt) \hat{\jmath} + p \hat{k}\)

where \(b,c,q,n,p\) are all constants. If the balloon is released from rest at the origin, what is the position vector of the balloon as a function of time? Give the answer in \(\hat{\imath} \; \hat{\jmath} \; \hat{k}\) notation.

Since the origin is the point of release, \(\vec{r_0}= 0\).

Since it is released from rest, \(\vec{v_0}=
0\).

\(\displaystyle \vec{v}(t) = \vec{v_0 }+ \int_0^t \vec{a}(t) dt = (bt - \frac{1}{3} ct^3) \hat{\imath }+ (qt + \frac{1}{2} nt^2) \hat{\jmath} + pt \hat{k}\)

\(\displaystyle \vec{r}(t) = \vec{r_0}+ \int_0^t \vec{v}(t) dt = (\frac{1}{2} b t^2 - \frac{1}{12} ct^4) \hat{\imath }+ (\frac{1}{2} qt^2 + \frac{1}{6} nt^3) \hat{\jmath} +\frac{1}{2} p t^2 \hat{k}\)

A basketball player throws the ball into the basket, as shown below. The basket is \(3.0\) m above the ground. The ball leaves his hands with a speed of \(10.0\) m/s at an angle of \(40.0\)° from horizontal, and it is in the air for \(1.2\) seconds.

Find the height above ground at which the ball is released (labeled \(y_0\) in the figure) and the horizontal distance the ball flies (labeled \(R\) in the figure).

Projectile motion equations are:

\[ \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} 0 \\ y_0 \end{bmatrix} +10 \begin{bmatrix} \cos 40° \\ \sin 40° \end{bmatrix} t +\frac{1}{2} \begin{bmatrix} 0 \\ -g \end{bmatrix} t^2 \]

At time \(t=1.2\) s:

\[ \begin{bmatrix} R \\ 3.0 \end{bmatrix} = \begin{bmatrix} 0 \\ y_0 \end{bmatrix} + \begin{bmatrix} 7.66 \\ 6.43 \end{bmatrix} (1.2) +\frac{1}{2} \begin{bmatrix} 0 \\ -9.8 \end{bmatrix} (1.2)^2 \]

\(x\)-motion gives \(R=9.19\) m.

\(y\)-motion gives \(y_0=2.34\) m.

*Last modified: Thu October 24 2024, 11:31 AM.*